## 4.3 Solve Mixture Applications with Systems of Equations

Learning objectives.

• Solve mixture applications
• Solve interest applications

## Solve applications of cost and revenue functions

Be prepared 4.3.

Before you get started, take this readiness quiz.

• Multiply: 4.025 ( 1,562 ) . 4.025 ( 1,562 ) . If you missed this problem, review Example 1.36 .
• Write 8.2% as a decimal. If you missed this problem, review Example 1.40 .

The ticket office at the zoo sold 553 tickets one day. The receipts totaled $3,936. How many$9 adult tickets and how many $6 child tickets were sold? ## Try It 4.48 The box office at a movie theater sold 147 tickets for the evening show, and receipts totaled$1,302. How many $11 adult and how many$8 child tickets were sold?

In the next example, we’ll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the ‘number’ column will be easy.

Greta wants to make 5 pounds of a nut mix using peanuts and cashews. Her budget requires the mixture to cost her $6 per pound. Peanuts are$4 per pound and cashews are $9 per pound. How many pounds of peanuts and how many pounds of cashews should she use? ## Try It 4.52 Sammy has most of the ingredients he needs to make a large batch of chili. The only items he lacks are beans and ground beef. He needs a total of 20 pounds combined of beans and ground beef and has a budget of$3 per pound. The price of beans is $1 per pound and the price of ground beef is$5 per pound. How many pounds of beans and how many pounds of ground beef should he purchase?

Another application of mixture problems relates to concentrated cleaning supplies, other chemicals, and mixed drinks. The concentration is given as a percent. For example, a 20% concentrated household cleanser means that 20% of the total amount is cleanser, and the rest is water. To make 35 ounces of a 20% concentration, you mix 7 ounces (20% of 35) of the cleanser with 28 ounces of water.

For these kinds of mixture problems, we’ll use “percent” instead of “value” for one of the columns in our table.

## Example 4.27

Sasheena is lab assistant at her community college. She needs to make 200 milliliters of a 40% solution of sulfuric acid for a lab experiment. The lab has only 25% and 50% solutions in the storeroom. How much should she mix of the 25% and the 50% solutions to make the 40% solution?

## Try It 4.53

LeBron needs 150 milliliters of a 30% solution of sulfuric acid for a lab experiment but only has access to a 25% and a 50% solution. How much of the 25% and how much of the 50% solution should he mix to make the 30% solution?

## Try It 4.54

Anatole needs to make 250 milliliters of a 25% solution of hydrochloric acid for a lab experiment. The lab only has a 10% solution and a 40% solution in the storeroom. How much of the 10% and how much of the 40% solutions should he mix to make the 25% solution?

Solve Interest Applications

The formula to model simple interest applications is I = P r t . I = P r t . Interest, I , is the product of the principal, P , the rate, r , and the time, t . In our work here, we will calculate the interest earned in one year, so t will be 1.

We modify the column titles in the mixture table to show the formula for interest, as you’ll see in the next example.

## Try It 4.57

Laura owes $18,000 on her student loans. The interest rate on the bank loan is 2.5% and the interest rate on the federal loan is 6.9%. The total amount of interest she paid last year was$1,066. What was the principal for each loan?

## Try It 4.58

Jill’s Sandwich Shoppe owes $65,200 on two business loans, one at 4.5% interest and the other at 7.2% interest. The total amount of interest owed last year was$3,582. What was the principal for each loan?

Suppose a company makes and sells x units of a product. The cost to the company is the total costs to produce x units. This is the cost to manufacture for each unit times x , the number of units manufactured, plus the fixed costs.

The revenue is the money the company brings in as a result of selling x units. This is the selling price of each unit times the number of units sold.

When the costs equal the revenue we say the business has reached the break-even point .

## Cost and Revenue Functions

The cost function is the cost to manufacture each unit times x , the number of units manufactured, plus the fixed costs.

The revenue function is the selling price of each unit times x , the number of units sold.

The break-even point is when the revenue equals the costs.

## Example 4.30

The manufacturer of a weight training bench spends $105 to build each bench and sells them for$245. The manufacturer also has fixed costs each month of $7,000. ⓐ Find the cost function C when x benches are manufactured. ⓑ Find the revenue function R when x benches are sold. ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. ⓓ Find the break-even point. Interpret what the break-even point means. ⓐ The manufacturer has$7,000 of fixed costs no matter how many weight training benches it produces. In addition to the fixed costs, the manufacturer also spends $105 to produce each bench. Suppose x benches are sold. Write the general Cost function formula. C ( x ) = ( cost per unit ) · x + fixed costs Substitute in the cost values. C ( x ) = 105 x + 7000 Write the general Cost function formula. C ( x ) = ( cost per unit ) · x + fixed costs Substitute in the cost values. C ( x ) = 105 x + 7000 ⓑ The manufacturer sells each weight training bench for$245. We get the total revenue by multiplying the revenue per unit times the number of units sold.

Write the general Revenue function. R ( x ) = ( selling price per unit ) · x Substitute in the revenue per unit. R ( x ) = 245 x Write the general Revenue function. R ( x ) = ( selling price per unit ) · x Substitute in the revenue per unit. R ( x ) = 245 x

ⓒ Essentially we have a system of linear equations. We will show the graph of the system as this helps make the idea of a break-even point more visual.

ⓓ To find the actual value, we remember the break-even point occurs when costs equal revenue.

Write the break-even formula. C ( x ) = R ( x ) 105 x + 7000 = 245 x Solve. 7000 = 140 x 50 = x Write the break-even formula. C ( x ) = R ( x ) 105 x + 7000 = 245 x Solve. 7000 = 140 x 50 = x

When 50 benches are sold, the costs equal the revenue.

The manufacturer of a weight training bench spends $120 to build each bench and sells them for$170. The manufacturer also has fixed costs each month of $150,000. Access this online resource for additional instruction and practice with interest and mixtures. • Interest and Mixtures ## Section 4.3 Exercises Practice makes perfect. In the following exercises, translate to a system of equations and solve. Tickets to a Broadway show cost$35 for adults and $15 for children. The total receipts for 1650 tickets at one performance were$47,150. How many adult and how many child tickets were sold?

Tickets for the Cirque du Soleil show are $70 for adults and$50 for children. One evening performance had a total of 300 tickets sold and the receipts totaled $17,200. How many adult and how many child tickets were sold? Tickets for an Amtrak train cost$10 for children and $22 for adults. Josie paid$1200 for a total of 72 tickets. How many children tickets and how many adult tickets did Josie buy?

Tickets for a Minnesota Twins baseball game are $69 for Main Level seats and$39 for Terrace Level seats. A group of sixteen friends went to the game and spent a total of $804 for the tickets. How many of Main Level and how many Terrace Level tickets did they buy? Tickets for a dance recital cost$15 for adults and $7 dollars for children. The dance company sold 253 tickets and the total receipts were$2771. How many adult tickets and how many child tickets were sold?

Tickets for the community fair cost $12 for adults and$5 dollars for children. On the first day of the fair, 312 tickets were sold for a total of $2204. How many adult tickets and how many child tickets were sold? Brandon has a cup of quarters and dimes with a total value of$ 3.80 . $3.80 . The number of quarters is four less than twice the number of quarters. How many quarters and how many dimes does Brandon have? Sherri saves nickels and dimes in a coin purse for her daughter. The total value of the coins in the purse is$ 0.95 . $0.95 . The number of nickels is two less than five times the number of dimes. How many nickels and how many dimes are in the coin purse? Peter has been saving his loose change for several days. When he counted his quarters and nickels, he found they had a total value$ 13.10 . $13.10 . The number of quarters was fifteen more than three times the number of dimes. How many quarters and how many dimes did Peter have? Lucinda had a pocketful of dimes and quarters with a value of$ 6.20 . $6.20 . The number of dimes is eighteen more than three times the number of quarters. How many dimes and how many quarters does Lucinda have? A cashier has 30 bills, all of which are$10 or $20 bills. The total value of the money is$460. How many of each type of bill does the cashier have?

A cashier has 54 bills, all of which are $10 or$20 bills. The total value of the money is $910. How many of each type of bill does the cashier have? Marissa wants to blend candy selling for$ 1.80 $1.80 per pound with candy costing$ 1.20 $1.20 per pound to get a mixture that costs her$ 1.40 $1.40 per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use? How many pounds of nuts selling for$6 per pound and raisins selling for $3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him$5 per pound?

Hannah has to make twenty-five gallons of punch for a potluck. The punch is made of soda and fruit drink. The cost of the soda is $1.79$ 1.79 per gallon and the cost of the fruit drink is $2.49$ 2.49 per gallon. Hannah’s budget requires that the punch cost $2.21$ 2.21 per gallon. How many gallons of soda and how many gallons of fruit drink does she need?

Joseph would like to make twelve pounds of a coffee blend at a cost of $6 per pound. He blends Ground Chicory at$5 a pound with Jamaican Blue Mountain at $9 per pound. How much of each type of coffee should he use? Julia and her husband own a coffee shop. They experimented with mixing a City Roast Columbian coffee that cost$7.80 per pound with French Roast Columbian coffee that cost $8.10 per pound to make a twenty-pound blend. Their blend should cost them$7.92 per pound. How much of each type of coffee should they buy?

Twelve-year old Melody wants to sell bags of mixed candy at her lemonade stand. She will mix M&M’s that cost $4.89 per bag and Reese’s Pieces that cost$3.79 per bag to get a total of twenty-five bags of mixed candy. Melody wants the bags of mixed candy to cost her $4.23 a bag to make. How many bags of M&M’s and how many bags of Reese’s Pieces should she use? Jotham needs 70 liters of a 50% solution of an alcohol solution. He has a 30% and an 80% solution available. How many liters of the 30% and how many liters of the 80% solutions should he mix to make the 50% solution? Joy is preparing 15 liters of a 25% saline solution. She only has 40% and 10% solution in her lab. How many liters of the 40% and how many liters of the 10% should she mix to make the 25% solution? A scientist needs 65 liters of a 15% alcohol solution. She has available a 25% and a 12% solution. How many liters of the 25% and how many liters of the 12% solutions should she mix to make the 15% solution? A scientist needs 120 milliliters of a 20% acid solution for an experiment. The lab has available a 25% and a 10% solution. How many liters of the 25% and how many liters of the 10% solutions should the scientist mix to make the 20% solution? A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 50% solution. How many liters of the 40% and how many liters of the 70% solutions will be used? A 90% antifreeze solution is to be mixed with a 75% antifreeze solution to get 360 liters of an 85% solution. How many liters of the 90% and how many liters of the 75% solutions will be used? Hattie had$3000 to invest and wants to earn 10.6 % 10.6 % interest per year. She will put some of the money into an account that earns 12% per year and the rest into an account that earns 10% per year. How much money should she put into each account?

Carol invested $2560 into two accounts. One account paid 8% interest and the other paid 6% interest. She earned 7.25 % 7.25 % interest on the total investment. How much money did she put in each account? Sam invested$48,000, some at 6% interest and the rest at 10%. How much did he invest at each rate if he received $4000 in interest in one year? Arnold invested$64,000, some at 5.5 % 5.5 % interest and the rest at 9%. How much did he invest at each rate if he received $4500 in interest in one year? After four years in college, Josie owes$65, 800 in student loans. The interest rate on the federal loans is 4.5 % 4.5 % and the rate on the private bank loans is 2%. The total interest she owes for one year was $2878.50 .$ 2878.50 . What is the amount of each loan?

Mark wants to invest $10,000 to pay for his daughter’s wedding next year. He will invest some of the money in a short term CD that pays 12% interest and the rest in a money market savings account that pays 5% interest. How much should he invest at each rate if he wants to earn$1095 in interest in one year?

A trust fund worth $25,000 is invested in two different portfolios. This year, one portfolio is expected to earn 5.25 % 5.25 % interest and the other is expected to earn 4%. Plans are for the total interest on the fund to be$1150 in one year. How much money should be invested at each rate?

A business has two loans totaling $85,000. One loan has a rate of 6% and the other has a rate of 4.5% This year, the business expects to pay$4,650 in interest on the two loans. How much is each loan?

Solve Applications of Cost and Revenue Functions

The manufacturer of an energy drink spends $1.20 to make each drink and sells them for$2. The manufacturer also has fixed costs each month of $8,000. ⓐ Find the cost function C when x energy drinks are manufactured. ⓑ Find the revenue function R when x drinks are sold. The manufacturer of a water bottle spends$5 to build each bottle and sells them for $10. The manufacturer also has fixed costs each month of$6500. ⓐ Find the cost function C when x bottles are manufactured. ⓑ Find the revenue function R when x bottles are sold. ⓒ Show the break-even point by graphing both the Revenue and Cost functions on the same grid. ⓓ Find the break-even point. Interpret what the break-even point means.

## Writing Exercises

Take a handful of two types of coins, and write a problem similar to Example 4.25 relating the total number of coins and their total value. Set up a system of equations to describe your situation and then solve it.

In Example 4.28 , we used elimination to solve the system of equations { s + b = 40,000 0.08 s + 0.03 b = 0.071 ( 40,000 ) . { s + b = 40,000 0.08 s + 0.03 b = 0.071 ( 40,000 ) .

Could you have used substitution or elimination to solve this system? Why?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Chapter 6: Polynomials

## 6.8 Mixture and Solution Word Problems

Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems. The basic structure of this table is shown below:

The first column in the table (Name) is used to identify the fluids or objects being mixed in the problem. The second column (Amount) identifies the amounts of each of the fluids or objects. The third column (Value) is used for the value of each object or the percentage of concentration of each fluid. The last column (Equation) contains the product of the Amount times the Value or Concentration.

Example 6.8.1

Jasnah has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Find the equation.

• The solution names are 50% (S 50 ), 60% (S 60 ), and 80% (S 80 ).
• The amounts are S 50 = 70 mL, S 80 , and S 60 = 70 mL + S 80 .
• The concentrations are S 50 = 0.50, S 60 = 0.60, and S 80 = 0.80.

The equation derived from this data is 0.50 (70 mL) + 0.80 (S 80 ) = 0.60 (70 mL + S 80 ).

Example 6.8.2

Sally and Terry blended a coffee mix that sells for $\2.50$ by mixing two types of coffee. If they used 40 mL of a coffee that costs $\3.00,$ how much of another coffee costing $\1.50$ did they mix with the first?

The equation derived from this data is:

$\begin{array}{ccccccc} 1.50(C_{1.50})&+&3.00(40)&=&2.50(40&+&C_{1.50}) \\ 1.50(C_{1.50})&+&120&=&100&+&2.50(C_{1.50}) \\ -2.50(C_{1.50})&-&120&=&-120&-&2.50(C_{1.50}) \\ \hline &&-1.00(C_{1.50})&=&-20&& \\ \\ &&(C_{1.50})&=&\dfrac{-20}{-1}&& \\ \\ &&C_{1.50}&=&20&& \end{array}$

This means 20 mL of the coffee selling for $\1.50$ is needed for the mix.

Example 6.8.3

Nick and Chloe have two grades of milk from their small dairy herd: one that is 24% butterfat and another that is 18% butterfat. How much of each should they use to end up with 42 litres of 20% butterfat?

The equation derived from this data is:

$\begin{array}{rrrrrrr} 0.24(B_{24})&+&0.18(42&- &B_{24})&=&0.20(42) \\ 0.24(B_{24})&+&7.56&-&0.18(B_{24})&=&8.4 \\ &-&7.56&&&&-7.56 \\ \hline &&&&0.06(B_{24})&=&0.84 \\ \\ &&&&B_{24}&=&\dfrac{0.84}{0.06} \\ \\ &&&&B_{24}&=&14 \end{array}$

This means 14 litres of the 24% buttermilk, and 28 litres of the 18% buttermilk is needed.

Example 6.8.4

In Natasha’s candy shop, chocolate, which sells for $\4$ a kilogram, is mixed with nuts, which are sold for $\2.50$ a kilogram. Chocolate and nuts are combined to form a chocolate-nut candy, which sells for $\3.50$ a kilogram. How much of each are used to make 30 kilograms of the mixture?

$\begin{array}{rrrrrrl} 4.00(C)&+&2.50(30&-&C)&=&3.50(30) \\ 4.00(C)&+&75&-&2.50(C)&=&105 \\ &-&75&&&&-75 \\ \hline &&&&1.50(C)&=&30 \\ \\ &&&&C&=&\dfrac{30}{1.50} \\ \\ &&&&C&=&20 \end{array}$

Therefore, 20 kg of chocolate is needed for the mixture.

With mixture problems, there is often mixing with a pure solution or using water, which contains none of the chemical of interest. For pure solutions, the concentration is 100%. For water, the concentration is 0%. This is shown in the following example.

Example 6.8.5

Joey is making a a 65% antifreeze solution using pure antifreeze mixed with water. How much of each should be used to make 70 litres?

$\begin{array}{rrrrl} 1.00(A)&+&0.00(70-A)&=&0.65(0.70) \\ &&1.00A&=&45.5 \\ &&A&=&45.5 \\ \end{array}$

This means the amount of water added is 70 L − 45.5 L = 24.5 L.

For questions 1 to 9, write the equations that define the relationship.

• A tank contains 8000 litres of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid?
• How much pure antifreeze should be added to 5 litres of a 30% mixture of antifreeze to make a solution that is 50% antifreeze?
• You have 12 kilograms of 10% saline solution and another solution of 3% strength. How many kilograms of the second should be added to the first in order to get a 5% solution?
• How much pure alcohol must be added to 24 litres of a 14% solution of alcohol in order to produce a 20% solution?
• How many litres of a blue dye that costs $\1.60$ per litre must be mixed with 18 litres of magenta dye that costs $\2.50$ per litre to make a mixture that costs $\1.90$ per litre?
• How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid?
• A 100-kg bag of animal feed is 40% oats. How many kilograms of pure oats must be added to this feed to produce a blend of 50% oats?
• A 20-gram alloy of platinum that costs $\220$ per gram is mixed with an alloy that costs $\400$ per gram. How many grams of the $\400$ alloy should be used to make an alloy that costs $\300$ per gram?
• How many kilograms of tea that cost $\4.20$ per kilogram must be mixed with 12 kilograms of tea that cost $\2.25$ per kilogram to make a mixture that costs $\3.40$ per kilogram?

Solve questions 10 to 21.

• How many litres of a solvent that costs $\80$ per litre must be mixed with 6 litres of a solvent that costs $\25$ per litre to make a solvent that costs $\36$ per litre?
• How many kilograms of hard candy that cost $\7.50$ per kg must be mixed with 24 kg of jelly beans that cost $\3.25$ per kg to make a mixture that sells for $\4.50$ per kg?
• How many kilograms of soil supplement that costs $\7.00$ per kg must be mixed with 20 kg of aluminum nitrate that costs $\3.50$ per kg to make a fertilizer that costs $\4.50$ per kg?
• A candy mix sells for $\2.20$ per kg. It contains chocolates worth $\1.80$ per kg and other candy worth $\3.00$ per kg. How much of each are in 15 kg of the mixture?
• A certain grade of milk contains 10% butterfat and a certain grade of cream 60% butterfat. How many litres of each must be taken so as to obtain a mixture of 100 litres that will be 45% butterfat?
• Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100 cc of a solution that is 68% acid?
• A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many litres of each must be used to make 600 litres of paint that is 19% green dye?
• How many kilograms of coffee that is 40% java beans must be mixed with coffee that is 30% java beans to make an 80-kg coffee blend that is 32% java beans?
• A caterer needs to make a slightly alcoholic fruit punch that has a strength of 6% alcohol. How many litres of fruit juice must be added to 3.75 litres of 40% alcohol?
• A mechanic needs to dilute a 70% antifreeze solution to make 20 litres of 18% strength. How many litres of water must be added?
• How many millilitres of water must be added to 50 millilitres of 100% acid to make a 40% solution?
• How many litres of water need to be evaporated from 50 litres of a 12% salt solution to produce a 15% salt solution?

## Module 5: Systems of Linear Equations

5.2 – applications of systems of linear equations, learning objectives.

• Specify what the variables in a cost/ revenue system of linear equations represent
• Determine and apply an appropriate method for solving the system

## (5.2.2) – Solve value problems with a system of linear equations

(5.2.3) – solve mixture problems with a system of linear equations, (5.2.4) – solve uniform motion problems with a system of linear equations.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible?

(credit: Thomas Sørenes)

## (5.2.1) – Solve cost and revenue problems

Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation $R=xp$, where $x=$ quantity and $p=$ price. The revenue function is shown in orange in the graph below.

The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The $x$ -axis represents quantity in hundreds of units. The y -axis represents either cost or revenue in hundreds of dollars.

The point at which the two lines intersect is called the break-even point . We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also$3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as $P\left(x\right)=R\left(x\right)-C\left(x\right)$. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

A business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of $y=0.85x+35,000$ and a revenue equation of $y=1.55x$:

• Interpret x and y for the cost equation
• Interpret x and y for the revenue equation

Cost: $y=0.85x+35,000$

Revenue:$y=1.55x$

The cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).

The revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them.  Let’s organize this information in a table:

## Example: Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function $C\left(x\right)=0.85x+35,000$ and the revenue function $R\left(x\right)=1.55x$, find the break-even point and the profit function.

Write the system of equations using $y$ to replace function notation.

$\begin{array}{l}\begin{array}{l}\\ y=0.85x+35,000\end{array}\hfill \\ y=1.55x\hfill \end{array}$

Substitute the expression $0.85x+35,000$ from the first equation into the second equation and solve for $x$.

$\begin{array}{c}0.85x+35,000=1.55x\\ 35,000=0.7x\\ 50,000=x\end{array}$

Then, we substitute $x=50,000$ into either the cost function or the revenue function. $1.55\left(50,000\right)=77,500$

The break-even point is $\left(50,000,77,500\right)$.

The profit function is found using the formula $P\left(x\right)=R\left(x\right)-C\left(x\right)$.

$\begin{array}{l}P\left(x\right)=1.55x-\left(0.85x+35,000\right)\hfill \\ \text{ }=0.7x - 35,000\hfill \end{array}$

The profit function is $P\left(x\right)=0.7x - 35,000$.

## Analysis of the Solution

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also$77,500. To make a profit, the business must produce and sell more than 50,000 units.

We see from the graph below that the profit function has a negative value until $x=50,000$, when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.

## How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

1) Identify unknown quantities in a problem represent them with variables.

2) Write a system of equations which models the problem’s conditions.

3) Solve the system.

4) Check proposed solution.

Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.

## Example: Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $25.00 for children and$50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets? Let c = the number of children and a = the number of adults in attendance. The total number of people is $2,000$. We can use this to write an equation for the number of people at the circus that day. $c+a=2,000$ The revenue from all children can be found by multiplying$25.00 by the number of children, $25c$. The revenue from all adults can be found by multiplying $50.00 by the number of adults, $50a$. The total revenue is$70,000. We can use this to write an equation for the revenue.

$25c+50a=70,000$

We now have a system of linear equations in two variables.

$\begin{array}{c}c+a=2,000\\ 25c+50a=70,000\end{array}$

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either $c$ or $a$. We will solve for $a$.

$\begin{array}{c}c+a=2,000\\ a=2,000-c\end{array}$

Substitute the expression $2,000-c$ in the second equation for $a$ and solve for $c$.

$\begin{array}{l} 25c+50\left(2,000-c\right)=70,000\hfill \\ 25c+100,000 - 50c=70,000\hfill \\ \text{ }-25c=-30,000\hfill \\ \text{ }c=1,200\hfill \end{array}$

Substitute $c=1,200$ into the first equation to solve for $a$.

$\begin{array}{l}1,200+a=2,000\hfill \\ \text{ }\text{}a=800\hfill \end{array}$

We find that $1,200$ children and $800$ adults bought tickets to the circus that day.

In this video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.

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## 17.1: Second-Order Linear Equations

• Last updated
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• Page ID 154053

• Gilbert Strang & Edwin “Jed” Herman

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## Learning Objectives

• Recognize homogeneous and nonhomogeneous linear differential equations.
• Determine the characteristic equation of a homogeneous linear equation.
• Use the roots of the characteristic equation to find the solution to a homogeneous linear equation.
• Solve initial-value and boundary-value problems involving linear differential equations.

When working with differential equations, usually the goal is to find a solution. In other words, we want to find a function (or functions) that satisfies the differential equation. The technique we use to find these solutions varies, depending on the form of the differential equation with which we are working. Second-order differential equations have several important characteristics that can help us determine which solution method to use. In this section, we examine some of these characteristics and the associated terminology.

## Homogeneous Linear Equations

Consider the second-order differential equation

$xy''+2x^2y'+5x^3y=0.\nonumber$

Notice that $$y$$ and its derivatives appear in a relatively simple form. They are multiplied by functions of $$x$$, but are not raised to any powers themselves, nor are they multiplied together. As discussed in previously, first-order equations with similar characteristics are said to be linear. The same is true of second-order equations. Also note that all the terms in this differential equation involve either $$y$$ or one of its derivatives. There are no terms involving only functions of $$x$$. Equations like this, in which every term contains $$y$$ or one of its derivatives, are called homogeneous .

Not all differential equations are homogeneous. Consider the differential equation

$xy''+2x^2y'+5x^3y=x^2.\nonumber$

The $$x^2$$ term on the right side of the equal sign does not contain $$y$$ or any of its derivatives. Therefore, this differential equation is nonhomogeneous .

## Definition: Homogeneous and Nonhomogeneous Linear Equations

A second-order differential equation is linear if it can be written in the form

$a_{2}(x)y''+a){1}(x)y'+a_{0}(x)y=r(x), \label{17.1}$

where $$a_{2}(x), a_{1}(x), a_{0}(x),$$ and $$r(x)$$ are real-valued functions and $$a_{2}(x)$$ is not identically zero. If $$r(x) \equiv 0$$—in other words, if $$r(x)=0$$ for every value of $$x$$—the equation is said to be a homogeneous linear equation . If $$r(x) \neq 0$$ for some value of $$x,$$ the equation is said to be a nonhomogeneous linear equation .

In linear differential equations, $$y$$ and its derivatives can be raised only to the first power and they may not be multiplied by one another. Terms involving $$y^2$$ or $$\sqrt{y'}$$ make the equation nonlinear. Functions of $$y$$ and its derivatives, such as $$\sin y$$ or $$e^{y'}$$, are similarly prohibited in linear differential equations.

Note that equations may not always be given in standard form (the form shown in the definition). It can be helpful to rewrite them in that form to decide whether they are linear, or whether a linear equation is homogeneous.

## Example $$\PageIndex{1}$$: Classifying Second-Order Equations

Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.

• $$y''+3x^4y'+x^2y^2=x^3$$
• $$(\sin x)y''+(\cos x)y'+3y=0$$
• $$4t^2x''+3txx'+4x=0$$
• $$5y''+y=4x^5$$
• $$( \cos x)y''- \sin y'+( \sin x)y- \cos x=0$$
• $$8ty''-6t^2y'+4ty-3t^2=0$$
• $$\sin(x^2)y''-( \cos x)y'+x^2y=y'-3$$
• $$y''+5xy'-3y= \cos y$$
• This equation is nonlinear because of the $$y^2$$ term.
• This equation is linear. There is no term involving a power or function of $$y,$$ and the coefficients are all functions of $$x$$.The equation is already written in standard form, and $$r(x)$$ is identically zero, so the equation is homogeneous.
• This equation is nonlinear. Note that, in this case, $$x$$ is the dependent variable and $$t$$is the independent variable. The second term involves the product of $$x$$ and $$x'$$, so the equation is nonlinear.
• This equation is linear. Since $$r(x)=4x^5,$$ the equation is nonhomogeneous.
• This equation is nonlinear, because of the $$\sin y'$$ term.

$8t^2y''-6t^2y'+4ty=3t^2. \nonumber$

$\sin(x^2)y''-(\cos x+1)y'+x^2y=-3. \nonumber$

• This equation is nonlinear because of the $$\cos y$$ term.

## Exercise $$\PageIndex{1}$$

• $$(y'')2-y'+8x^3y=0$$
• $$(\sin t)y''+ \cos t-3ty'=0$$

Write the equation in standard form (Equation \ref{17.1}) if necessary. Check for powers or functions of $$y$$ and its derivatives.

Nonlinear Linear

nonhomogeneous

Later in this section, we will see some techniques for solving specific types of differential equations. Before we get to that, however, let’s get a feel for how solutions to linear differential equations behave. In many cases, solving differential equations depends on making educated guesses about what the solution might look like. Knowing how various types of solutions behave will be helpful.

## Example $$\PageIndex{2}$$: Verifying a Solution

Consider the linear, homogeneous differential equation

$x^2y''-xy′-3y=0. \nonumber$

Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of $$x$$ associated with higher-order derivatives of $$y$$. Show that $$y=x^3$$ is a solution to this differential equation.

Let $$y=x^3.$$ Then $$y'=3x^2$$ and $$y''=6x.$$ Substituting into the differential equation, we see that

\begin{align*} x^2y''-xy'-3y &=x^2(6x)-x(3x^2)-3(x^3) \\[4pt] &=6x^3-3x^3-3x^3 \\[4pt] &=0. \end{align*}

## Exercise $$\PageIndex{2}$$

Show that $$y=2x^2$$ is a solution to the differential equation

$\dfrac{1}{2}x^2y''-xy'+y=0. \label{ex2}$

Calculate the derivatives and substitute them into the differential equation.

This requires calculating $$y'$$ and $$y''$$.

$y' = \dfrac{dy}{dx} = 4x \nonumber$

$y'' = \dfrac{dy'}{dx} = 4 \nonumber$

Inserting these derivatives along with $$y=2x^2$$ into Equation \ref{ex2}.

\begin{align*} \dfrac{1}{2}x^2y''-xy'+y &\overset{?}{=} 0 \\[4pt] \dfrac{1}{2}x^2(4) - x (4x) + 2x^2 &\overset{?}{=} 0 \\[4pt] 2x^2 - 4x^2 + 2x^2 &\overset{\checkmark}{=} 0 \end{align*} \nonumber

Yes, this is a solution to the differential equation in Equation \ref{ex2}.

Although simply finding any solution to a differential equation is important, mathematicians and engineers often want to go beyond finding one solution to a differential equation to finding all solutions to a differential equation. In other words, we want to find a general solution . Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation. An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact as the following theorem.

## Theorem: SUPERPOSITION PRINCIPLE

If $$y_1(x)$$ and $$y_2(x)$$ are solutions to a linear homogeneous differential equation, then the function

$y(x)=c_1y_1(x)+c_2y_2(x), \label{super}$

where $$c_1$$ and $$c_2$$ are constants, is also a solution.

The proof of this superposition principle theorem is left as an exercise.

## Example $$\PageIndex{3}$$: Verifying the Superposition Principle

Consider the differential equation

$y''-4y'-5y=0.\nonumber$

Given that $$e^{-x}$$ and $$e^{5x}$$ are solutions to this differential equation, show that $$4e^{-x}+e^{5x}$$ is a solution.

Although this can be done through a simple application of the Superposition principle (Equation \ref{super}), but we can also confirm it is a solution via an approach like in Example $$\PageIndex{2}$$. We have

\begin{align*} y(x) &=4e^{-x}+e^{5x} \\[4pt] y'(x) &= -4e^{-x} + 5e^{5x} \\[4pt] y''(x) &=4e^{-x}+25e^{5x}. \end{align*}

\begin{align*} y''-4y'-5y &\overset{?}{=} (4e^{-x}+25e^{5x})-4(-4e^{-x}+5e^{5x})-5(4e^{-x}+e^{5x}) \\[4pt] &\overset{?}{=} 4e^{-x}+25e^{5x}+16e^{-x}-20e^{5x}-20e^{-x}-5e^{5x} \\[4pt] &\overset{\checkmark}{=}0. \end{align*} \nonumber

Thus, $$y(x)=4e^{-x}+e^{5x}$$ is a solution.

## Exercise $$\PageIndex{3}$$

$y''+5y'+6y=0. \nonumber$

Given that $$e^{-2x}$$ and $$e^{-3x}$$ are solutions to this differential equation, show that $$3e^{-2x}+6e^{-3x}$$ is a solution.

Differentiate the function and substitute into the differential equation.

Although this can be a simple application of the Superposition principle (Equation \ref{super}), we can also set through it like in Example $$\PageIndex{2}$$. We have

\begin{align*} y(x) &=3e^{-2x}+6e^{-3x} \\[4pt] y'(x) &= -6 e^{-2x} - 18e^{-3x} \\[4pt] y''(x) &= 12e^{-2x} + 54e^{3x}. \end{align*}

\begin{align*} y''+5y'+6y &= (12e^{-2x} + 54e^{3x}) + 5( -6 e^{-2x} - 18e^{-3x} ) + 6( 3e^{-2x} + 6e^{3x}) \\[4pt] &\overset{?}{=} \cancel{12e^{-2x}} + \bcancel{54e^{3x}} - \cancel{30e^{-2x}} - \bcancel{90e^{3x}} + \cancel{18e^{-2x}} + \bcancel{36e^{3x}} \\[4pt] &\overset{\checkmark}{=}0. \end{align*} \nonumber

Thus, $$3e^{-2x}+6e^{-3x}$$ is a solution to the differential equation

Unfortunately, to find the general solution to a second-order differential equation, it is not enough to find any two solutions and then combine them. Consider the differential equation

$x''+7x'+12x=0.\nonumber$

Both $$e^{-3t}$$ and $$2e^{-3t}$$ are solutions (you can check this). However,

$x(t)=c_1e^{-3t}+c_2(2e^{-3t})\nonumber$

is not the general solution. This expression does not account for all solutions to the differential equation. In particular, it fails to account for the function $$e^{-4t},$$ which is also a solution to the differential equation. It turns out that to find the general solution to a second-order differential equation, we must find two linearly independent solutions . We define that terminology here.

## Definition: Linearly Dependent functions

A set of functions $$f_1(x),\, f_2(x), \ldots ,f_n(x)$$ is said to be linearly dependent if there are constants $$c_1,\, c_2, \ldots c_n,$$, not all zero, such that

$c_1f_1(x)+c_2f_2(x)+ \cdots +c_nf_n(x)=0 \nonumber$

for all $$x$$ over the interval of interest. A set of functions that is not linearly dependent is said to be linearly independent .

In this chapter, we usually test sets of only two functions for linear independence, which allows us to simplify this definition. From a practical perspective, we see that two functions are linearly dependent if either one of them is identically zero or if they are constant multiples of each other.

First we show that if the functions meet the conditions given previously, then they are linearly dependent. If one of the functions is identically zero—say, $$f_2(x) \equiv 0$$—then choose $$c_1=0$$ and $$c_2=1,$$ and the condition for linear dependence is satisfied. If, on the other hand, neither $$f_1(x)$$ nor $$f_2(x)$$ is identically zero, but $$f_1(x)=Cf_2(x)$$ for some constant $$C,$$ then choose $$c_1=C$$ and $$c_2=-1,$$ and again, the condition is satisfied.

Next, we show that if two functions are linearly dependent, then either one is identically zero or they are constant multiples of one another. Assume $$f_1(x)$$ and $$f_2(x)$$ are linearly independent. Then, there are constants, $$c_1$$ and $$c_2,$$ not both zero, such that

$c_1f_1(x)+c_2f_2(x)=0 \nonumber$

for all $$x$$ over the interval of interest. Then,

$c_1f_1(x)=-c_2f_2(x). \nonumber$

Now, since we stated that $$c_1$$ and $$c_2$$ can’t both be zero, assume $$c_2 \neq 0.$$ Then, there are two cases: either $$c_1=0$$ or $$c_1\neq 0.$$ If $$c_1=0,$$ then

\begin{align*} 0 &=-c_2f_2(x) \\[4pt] 0 &=f_2(x), \end{align*}

so one of the functions is identically zero. Now suppose $$c_1 \neq 0.$$ Then,

$f_1(x)=\left(- \dfrac{c_2}{c_1}\right)f_2(x) \nonumber$

and we see that the functions are constant multiples of one another.

## Theorem: Linear Dependence of Two Functions

Two functions, $$f_1(x)$$ and $$f_2(x),$$ are said to be linearly dependent if either one of them is identically zero or if $$f_1(x)=Cf_2(x)$$ for some constant $$C$$ and for all $$x$$ over the interval of interest. Functions that are not linearly dependent are said to be linearly independent .

## Example $$\PageIndex{4}$$: Testing for Linear Dependence

Determine whether the following pairs of functions are linearly dependent or linearly independent.

• $$f_1(x)=x^2$$ and $$f_2(x)=5x^2$$
• $$f_1(x)= \sin x$$ and $$f_2(x)= \cos x$$
• $$f_1(x)=e^{3x}$$ and $$f_2(x)=e^{-3x}$$
• $$f_1(x)=3x$$ and $$f_2(x)=3x+1$$
• $$f_2(x)=5f_1(x),$$ so the functions are linearly dependent.
• There is no constant $$C$$ such that $$f_1(x)=Cf_2(x),$$ so the functions are linearly independent.
• There is no constant $$C$$ such that $$f_1(x)=Cf_2(x),$$ so the functions are linearly independent. Don’t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.

## Exercise $$\PageIndex{4}$$

Determine whether the following pairs of functions are linearly dependent or linearly independent: $$f_1(x)=e^{x}$$ and $$f_2(x)=3e^{3x}.$$

Are the functions constant multiples of one another?

Linearly independent

If we are able to find two linearly independent solutions to a second-order differential equation, then we can combine them to find the general solution. This result is formally stated in the following theorem.

## Theorem: General Solution to a Homogeneous Equation

If $$y_1(x)$$ and $$y_2(x)$$ are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by

$y(x)=c_1y_1(x)+c_2y_2(x), \nonumber$

where $$c_1$$ and $$c_2$$ are constants.

When we say a family of functions is the general solution to a differential equation, we mean that

• every expression of that form is a solution and
• every solution to the differential equation can be written in that form, which makes this theorem extremely powerful.

If we can find two linearly independent solutions to a second order differential equation, we have, effectively, found all solutions to the second order differential equation—quite a remarkable statement. The proof of this theorem is beyond the scope of this text.

## Example $$\PageIndex{5}$$: Writing the General Solution

If $$y_1(t)=e^{3t}$$ and $$y_2(t)=e^{-3t}$$ are solutions to $$y''-9y=0,$$ what is the general solution?

Note that $$y_1$$ and $$y_2$$ are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is

$$y(t)=c_1e^{3t}+c_2e^{-3t}.$$

## Exercise $$\PageIndex{5}$$

If $$y_1(x)=e^{3x}$$ and $$y_2(x)=xe^{3x}$$ are solutions to $$y''-6y'+9y=0,$$ what is the general solution?

Check for linear independence first.

$$y(x)=c_1e^{3x}+c_2xe^{3x}$$

## Second-Order Equations with Constant Coefficients

Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form

$ay''+by'+cy=0, \tag{17.2}$

where $$a, b,$$ and $$c$$ are constants.

Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let’s see what happens when we try a solution of the form $$y(x)=e^{ \lambda x}$$, where $$\lambda$$ (the lowercase Greek letter lambda) is some constant.

If $$y(x)=e^{ \lambda x}$$, then $$y'(x)= \lambda e^{ \lambda x}$$ and $$y''= \lambda^2 e^{ \lambda x}.$$ Substituting these expressions into Equation \ref{17.1}, we get

\begin{align*} ay''+by'+cy &= a(\lambda^2e^{\lambda x})+b(\lambda e^{\lambda x})+ce^{\lambda x} \\[4pt] &=e^{\lambda x}(a \lambda^2+b \lambda +c). \end{align*}

Since $$e^{\lambda x}$$ is never zero, this expression can be equal to zero for all $$x$$ only if

$a\lambda^2+b\lambda +c=0. \nonumber$

We call this the characteristic equation of the differential equation.

## Definition: characteristic equation

The characteristic equation of the second order differential equation $$ay''+by'+cy=0$$ is

The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula

$\lambda = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}. \nonumber$

This gives three cases. The characteristic equation has

• distinct real roots;
• a single, repeated real root; or
• complex conjugate roots.

We consider each of these cases separately.

## Case 1: Distinct Real Roots

If the characteristic equation has distinct real roots $$\lambda_1$$ and $$\lambda_2$$, then $$e^{\lambda_1x}$$ and $$e^{\lambda_2x}$$ are linearly independent solutions to Example \ref{17.1}, and the general solution is given by

$y(x)=c_1e^{\lambda_1x}+c_2e^{\lambda_2x}, \nonumber$

For example, the differential equation $$y''+9y'+14y=0$$ has the associated characteristic equation $$\lambda^2+9\lambda+14=0.$$ This factors into $$(\lambda +2)(\lambda +7)=0,$$ which has roots $$\lambda_1=-2$$ and $$\lambda_2=-7.$$ Therefore, the general solution to this differential equation is

$y(x)=c_1e^{-2x}+c_2e^{-7x}. \nonumber$

## Case 2: Single Repeated Real Root

Things are a little more complicated if the characteristic equation has a repeated real root, $$\lambda$$. In this case, we know $$e^{\lambda x}$$ is a solution to Equation \ref{17.1}, but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form $$ke^{\lambda x},$$ where $$k$$ is some constant, but it would not be linearly independent of $$e^{\lambda x}.$$ Therefore, let’s try $$xe^{\lambda x}$$ as the second solution. First, note that by the quadratic formula,

$\lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a}. \nonumber$

But, $$\lambda$$ is a repeated root, so the discriminate ($$b^2-4ac$$) is zero and $$\lambda = \frac{-b}{2a}$$. Thus, if $$y=xe^{\lambda x}$$, we have

\begin{align*} y' =e^{\lambda x}+ \lambda xe^{\lambda x} \\[4pt] y'' =2\lambda e^{\lambda x}+\lambda^2xe^{\lambda x}. \end{align*}

Substituting both expressions into Equation \ref{17.1}, we see that

\begin{align*} ay''+by′+cy &=a(2λe^{λx}+λ^2xe^{λx})+b(e^{λx}+λxe^{λx})+cxe^{λx} \\[4pt] &=xe^{λx}(aλ^2+bλ+c)+e^{λx}(2aλ+b) \\[4pt] &=xe^{λx}(0)+e^{λx}(2a(−b2a)+b)\\[4pt] &=0+e^{λx}(0) \\[4pt] &\overset{\checkmark}{=}0. \end{align*}

This shows that $$xe^{\lambda x}$$ is a solution to Equation \ref{17.1}. Since $$e^{\lambda x}$$ and $$xe^{\lambda x}$$ are linearly independent, when the characteristic equation has a repeated root $$\lambda$$, the general solution to Equation \ref{17.1} is given by

$y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}, \nonumber$

For example, the differential equation $$y''+12y'+36y=0$$ has the associated characteristic equation

$\lambda^2+12 \lambda +36=0.\nonumber$

This factors into $$(\lambda +6)^2=0,$$ which has a repeated root $$\lambda =-6$$. Therefore, the general solution to this differential equation is

$y(x)=c_1e^{-6x}+c_2xe^{-6x}.\nonumber$

## Case 3: Complex Conjugate Roots

The third case we must consider is when $$b^2-4ac <0.$$ In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number $$i= \sqrt{-1}$$ to find the roots, which take the form $$\lambda_1= \alpha + \beta i$$ and $$\lambda _2=\alpha -\beta i.$$ The complex number $$\alpha +\beta i$$ is called the conjugate of $$\alpha -\beta i$$. Thus, we see that when the discriminate $$b^2-4ac$$ is negative, the roots of our characteristic equation are always complex conjugates .

This creates a little bit of a problem for us. If we follow the same process we used for distinct real roots—using the roots of the characteristic equation as the coefficients in the exponents of exponential functions—we get the functions $$e^{(\alpha + \beta i)x}$$ and $$e^{(\alpha - \beta i)x}$$ as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-value functions, in this course we do not address the idea of a derivative for such functions. So, if possible, we’d like to find two linearly independent real-value solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions $$e^{(\alpha + \beta i)x}$$ and $$e^{(\alpha - \beta i)x}$$ as if they were real-value functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-value functions do not follow the same rules as real-value functions. Those of you interested in a more in-depth discussion of complex-value functions should consult a complex analysis text.

Based on the roots $$\alpha \pm \beta i$$ of the characteristic equation, the functions $$e^{(\alpha + \beta i)x}$$ and $$e^{(\alpha - \beta i)x}$$ are linearly independent solutions to the differential equation and the general solution is given by

$y(x)=c_1e^{(\alpha +\beta i)x}+c_2e^{(\alpha - \beta i)x}. \nonumber$

Using some smart choices for $$c_1$$ and $$c_2$$, and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to Equation \ref{17.1} and express our general solution in those terms.

We encountered exponential functions with complex exponents earlier. One of the key tools we used to express these exponential functions in terms of sines and cosines was Euler’s formula, which tells us that

$\underbrace{e^{i \theta }= \cos \theta+ i \sin \theta}_{\text{Euler’s formula}} \label{Euler}$

for all real numbers $$\theta$$.

Going back to the general solution, we have

\begin{align*} y(x) &=c_1e^{( \alpha+ \beta i)x}+c_2e^{(\alpha - \beta i)x} \\[4pt] &=c_1e^{\alpha x}e^{\beta ix}+c_2e^{\alpha x}e^{- \beta ix} \\[4pt] &=e^{\alpha x}(c_1e^{\beta ix}+c_2e^{-\beta ix}).\end{align*}

Applying Euler’s formula (Equation \ref{Euler}) together with the identities $$\cos(-x)=\cos x$$ and $$\sin(-x)=- \sin x,$$ we get

\begin{align} y(x) &=e^{\alpha x}[c_1(\cos \beta x+i \sin \beta x)+c_2(\cos(- \beta x)+i \sin(- \beta x))] \nonumber \\[4pt] &=e^{\alpha x}[(c_1+c_2)\cos \beta x+(c_1-c_2)i \sin \beta x]. \label{E1}\end{align}

Now, if we choose $$c_1=c_2= \frac{1}{2},$$ the second term is zero and we get

$y(x)=e^{\alpha x} \cos \beta x \nonumber$

as a real-value solution to Equation \ref{17.1}. Similarly, if we choose $$c_1=−\frac{i}{2}$$ and $$c_2=\frac{i}{2}$$, the first term of Equation \ref{E1} is zero and we get

$y(x)=e^{\alpha x} \sin \beta x \nonumber$

as a second, linearly independent, real-value solution to Equation \ref{17.1}.

Based on this, we see that if the characteristic equation has complex conjugate roots $$\alpha \pm \beta i,$$ then the general solution to Equation \ref{17.1} is given by

\begin{align*} y(x) &=c_1e^{\alpha x} \cos \beta x+c_2e^{\alpha x} \sin \beta x \\[4pt] &=e^{\alpha x}(c_1 \cos \beta x+c_2 \sin \beta x),\end{align*}

For example, the differential equation $$y''-2y'+5y=0$$ has the associated characteristic equation $$\lambda ^2-2 \lambda +5=0.$$ By the quadratic formula, the roots of the characteristic equation are $$1\pm 2i.$$ Therefore, the general solution to this differential equation is

$y(x)=e^{x}(c_1 \cos 2x+c_2 \sin 2x).\nonumber$

## Summary of Results

We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots. The three cases are summarized in Table $$\PageIndex{1}$$.

## PROBLEM-SOLVING STRATEGY: USING THE CHARACTERISTIC EQUATION TO SOLVE SECOND-ORDER DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

• Write the differential equation in the form $$a''+by'+cy=0.$$
• Find the corresponding characteristic equation $$a\lambda^2+b\lambda +c=0.$$
• Either factor the characteristic equation or use the quadratic formula to find the roots.
• Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.

## Example $$\PageIndex{6}$$: Solving Second-Order Equations with Constant Coefficients

Find the general solution to the following differential equations. Give your answers as functions of $$x$$.

• $$y''+3y'-4y=0$$
• $$y''+6y'+13y=0$$
• $$y''+2y'+y=0$$
• $$y''-5y'=0$$
• $$y''-16y=0$$
• $$y''+16y=0$$

Note that all these equations are already given in standard form (step 1).

$y(x)=c_1e^{-4x}+c_2e^{x}. \tag{step 1}$

$y(t)=e^{-3t}(c_1 \cos 2t+c_2 \sin 2t). \tag{step 2}$

$y(t)=c_1e^{-t}+c_2te^{-t}. \tag{step 3}$

$y(x)=c_1+c_2e^{5x}. \tag{step 4}$

$y(x)=c_1e^{4x}+c_2e^{-4x}. \tag{step 5}$

$y(t)=c_1 \cos 4t+c_2 \sin 4t. \tag{step 6}$

## Exercise $$\PageIndex{6}$$

Find the general solution to the following differential equations:

• $$y''-2y'+10y=0$$
• $$y''+14y'+49y=0$$

Find the roots of the characteristic equation.

$$y(x)=e^x(c_1 \cos 3x+c_2 \sin 3x)$$

$$y(x)=c_1e^{-7x}+c_2xe^{-7x}$$

## Initial-Value Problems and Boundary-Value Problems

So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time. For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time $$t=t_0.$$ This means the system is at equilibrium, so $$y(t_0)=0,$$ and the compression of the shock absorber is not changing, so $$y'(t_0)=0.$$ With these two initial conditions and the general solution to the differential equation, we can find the specific solution to the differential equation that satisfies both initial conditions. This process is known as solving an initial-value problem . (Recall that we discussed initial-value problems in Introduction to Differential Equations.) Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.

Sometimes we know the condition of the system at two different times. For example, we might know $$y(t_0)=y_0$$ and $$y(t_1)=y_1.$$These conditions are called boundary conditions, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem.

Mathematicians, scientists, and engineers are interested in understanding the conditions under which an initial-value problem or a boundary-value problem has a unique solution. Although a complete treatment of this topic is beyond the scope of this text, it is useful to know that, within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.

## Example $$\PageIndex{7}$$: Solving an Initial-Value Problem

Solve the following initial-value problem: $$y''+3y'-4y=0, \, y(0)=1,\, y'(0)=-9.$$

We already solved this differential equation in Example 17.6a. and found the general solution to be

$y(x)=c_1e^{-4x}+c_2e^{x}. \nonumber$

$y'(x)=-4c_1e^{-4x}+c_2e^{x}. \nonumber$

When $$x=0,$$ we have $$y(0)=c_1+c_2$$ and $$y'(0)=-4c_1+c_2.$$ Applying the initial conditions, we have

\begin{align*} c_1+c_2 &=1 \\[4pt] -4c_1+c_2 &=-9.\end{align*}

Then $$c_1=1-c_2.$$ Substituting this expression into the second equation, we see that

\begin{align*} -4(1-c_2)+c_2 &= -9 \\[4pt] -4+4c_2+c_2 &=-9 \\[4pt] 5c_2 &=-5 \\[4pt] c_2 &=-1. \end{align*}

So, $$c_1=2$$ and the solution to the initial-value problem is

$y(x)=2e^{-4x}-e^{x}. \nonumber$

## Exercise $$\PageIndex{7}$$

Solve the initial-value problem $$y''-3y'-10y=0, \quad y(0)=0, \; y'(0)=7.$$

Use the initial conditions to determine values for $$c_1$$ and $$c_2$$.

$y(x)=-e^{-2x}+e^{5x} \nonumber$

## Example $$\PageIndex{8}$$: Solving an Initial-Value Problem and Graphing the Solution

Solve the following initial-value problem and graph the solution:

$y''+6y'+13y=0, \quad y(0)=0, \; y'(0)=2\nonumber$

We already solved this differential equation in Example $$\PageIndex{6b}$$. and found the general solution to be

$y(x)=e^{-3x}(c_1 \cos 2x+c_2 \sin 2x).\nonumber$

$y'(x)=e^{-3x}(-2c_1 \sin 2x+2c_2 \cos 2x)-3e^{-3x}(c_1 \cos 2x+c_2 \sin 2x). \nonumber$

When $$x=0,$$ we have $$y(0)=c_1$$ and $$y'(0)=2c_2-3c_1$$. Applying the initial conditions, we obtain

\begin{align*} c_1 &=0 \\[4pt] -3c_1+2c_2 &=2. \end{align*}

Therefore, $$c_1=0, \, c_2=1,$$ and the solution to the initial value problem is shown in the following graph.

$y=e^{-3x} \sin 2x.\nonumber$

## Exercise $$\PageIndex{8}$$

Solve the following initial-value problem and graph the solution: $$y''-2y'+10y=0, \quad y(0)=2, \; y'(0)=-1$$

Use the initial conditions to determine values for $$c_1$$ and $$c_2.$$

$y(x)=e^{x}(2 \cos 3x - \sin 3x) \nonumber$

## Example $$\PageIndex{9}$$: Initial-Value Problem Representing a Spring-Mass System

The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in Applications. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)

$y''+2y'+y=0, \quad y(0)=1, \; y'(0)=0 \nonumber$

Solve the initial-value problem and graph the solution. What is the position of the mass at time $$t=2$$ sec? How fast is the mass moving at time $$t=1$$ sec? In what direction?

In Example Example $$\PageIndex{6c}$$. we found the general solution to this differential equation to be

$y(t)=c_1e^{-t}+c_2te^{-t}. \nonumber$

$y'(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t}). \nonumber$

When $$t=0,$$ we have $$y(0)=c_1$$ and $$y'(0)=c_1+c_2.$$ Applying the initial conditions, we obtain

$c_1=1 \\ -c_1+c_2=0. \nonumber$

Thus, $$c_1=1, c_2=1,$$ and the solution to the initial value problem is

$y(t)=e^{-t}+te^{-t}. \nonumber$

This solution is represented in the following graph. At time $$t=2,$$ the mass is at position $$y(2)=e^{-2}+2e^{-2}=3e^{-2} \approx 0.406$$ m below equilibrium.

To calculate the velocity at time $$t=1,$$ we need to find the derivative. We have $$y(t)=e^{-t}+te^{-t},$$ so

$y'(t)=-e^{-t}+e^{-t}-te^{-t}= -te^{-t}. \nonumber$

Then $$y'(1)=-e^{-1} \approx -0.3679$$. At time $$t=1,$$ the mass is moving upward at $$0.3679$$ m/sec.

## Exercise $$\PageIndex{9}$$

Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time $$t=0.3$$ sec? How fast is it moving at time $$t=0.1$$ sec? In what direction?

$y''+14y'+49y=0, \quad y(0)=0, \; y'(0)=1 \nonumber$

$y(t)=te^{-7t}\nonumber$

At time $$t=0.3, \; y(0.3)=0.3e^{(-7^{\ast} 0.3)}=0.3e^{-2.1} \approx 0.0367.$$ The mass is $$0.0367$$ft below equilibrium. At time $$t=0.1, \; y'(0.1)=0.3e^{-0.7} \approx 0.1490.$$ The mass is moving downward at a speed of $$0.1490$$ ft/sec.

## Example $$\PageIndex{10}$$: Solving a Boundary-Value Problem

In Example 17.6f. we solved the differential equation $$y''+16y=0$$ and found the general solution to be $$y(t)=c_1 \cos 4t+c_2 \sin 4t.$$ If possible, solve the boundary-value problem if the boundary conditions are the following:

• $$y(0)=0, y( \frac{\pi}{4})=0$$
• $$y(0)=1,y(0)=1, y(\frac{\pi}{8})=0$$
• $$y(\frac{\pi}{8})=0, y(\frac{3 \pi}{8})=2$$

$y(x)=c_1 \cos 4t+c_2 \sin 4t. \nonumber$

• Applying the first boundary condition given here, we get $$y(0)=c_1=0.$$ So the solution is of the form $$y(t)=c_2 \sin 4t.$$ When we apply the second boundary condition, though, we get $$y(\frac{\pi}{4})=c_2 \sin(4(\frac{\pi}{4}))=c_2 \sin \pi =0$$ for all values of $$c_2$$. The boundary conditions are not sufficient to determine a value for $$c_2,$$ so this boundary-value problem has infinitely many solutions. Thus, $$y(t)=c_2 \sin 4t$$ is a solution for any value of $$c_2$$.
• Applying the first boundary condition given here, we get $$y(0)=c_1=1.$$ Applying the second boundary condition gives $$y(\frac{\pi}{8})=c_2=0,$$ so $$c_2=0.$$ In this case, we have a unique solution: $$y(t)= \cos 4t$$.
• Applying the first boundary condition given here, we get $$y(\frac{\pi}{8})=c_2=0.$$ However, applying the second boundary condition gives $$y(\frac{3 \pi}{8})=-c_2=2,$$ so $$c_2=-2.$$ We cannot have $$c_2=0=-2,$$ so this boundary value problem has no solution.

## Key Concepts

• Second-order differential equations can be classified as linear or nonlinear, homogeneous or nonhomogeneous.

$y(x)=c_1y_1(x)+c_2y_2(x).\nonumber$

• To solve homogeneous second-order differential equations with constant coefficients, find the roots of the characteristic equation. The form of the general solution varies depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.
• Initial conditions or boundary conditions can then be used to find the specific solution to a differential equation that satisfies those conditions, except when there is no solution or infinitely many solutions.

## Key Equations

• Linear second-order differential equation $a_2(x)y''+a_1(x)y'+a_0(x)y=r(x) \nonumber$
• Second-order equation with constant coefficients $ay''+by'+cy=0 \nonumber$

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26. 17.1: Second-Order Linear Equations

as a second, linearly independent, real-value solution to Equation 17.1.1. Based on this, we see that if the characteristic equation has complex conjugate roots α ± βi, then the general solution to Equation 17.1.1 is given by. y(x) = c1eαxcosβx + c2eαxsinβx = eαx(c1cosβx + c2sinβx), where c1 and c2 are constants.