# The value of r. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.4, Problem 66E
To determine

## To find: The value of r.

Expert Solution

The value of r=2±3.

### Explanation of Solution

Given:

The function y=erx.

The differential equation y4y+y=0.

Derivative rule:

Constant multiple rule: ddx(cf)=cddx(f)

Calculation:

Obtain the first derivative of y.

y(x)=ddx(erx)=rerx

Therefore, the first derivative of y is y=rerx.

Obtain the second derivative of y=erx.

y(x)=ddx(y)=ddx(rerx)

Apply the constant multiple rule (1),

y(x)=rddx(erx)=r(rerx)=r2erx

Therefore, the second derivative of y is y=r2erx_.

Substitute the values of y(x), y(x) and y(x) in y4y+y=0,

(r2erx)4(rerx)+(erx)=0erx(r24r+1)=0

Since erx=0 does not exit, r24r+1=0.

Use the quadratic formula r=b±b24ac2a with a=1, b=4 and c=1,

r=(4)±(4)24(1)(1)2(1)=4±1642=4±122=4±432

On further simplification to obtain the roots of the equation r24r+1=0.

r=4±232=2(2±3)2=2±3

Therefore, the value of r=2±3.

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