# The function y = e 2 x ( A cos 3 x + B sin 3 x ) satisfies the differential equation y ″ − 4 y ′ + 13 y = 0 .

### 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 65E
To determine

## To show: The function y=e2x(Acos3x+Bsin3x) satisfies the differential equation y″−4y′+13y=0.

Expert Solution

### Explanation of Solution

Derivative rules:

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

(2) Product rule: ddx(fg)=fddx(g)+gddx(f)

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

Proof:

Obtain the first derivative of y=e2x(Acos3x+Bsin3x).

y(x)=ddx(e2x(Acos3x+Bsin3x))

Apply the sum rule (3) and the constant multiple rule (1),

Apply the product rule (2),

y(x)=A(e2xddx(cos3x)+cos3xddx(e2x))+B(e2xddx(sin3x)+sin3xddx(e2x))=A(e2x(3(sin3x))+cos3x(2e2x))+B(e2x(cos3x3)+sin3x(2e2x))=e2x(3Asin3x+2Acos3x)+e2x(3Bcos3x+2Bsin3x)=e2x[(3A+2B)sin3x+(2A+3B)cos3x]

Therefore, the first derivative of y is y=e2x[(3A+2B)sin3x+(2A+3B)cos3x].

Obtain the second derivative of y=e2x(Acos3x+Bsin3x).

y(x)=ddx(dydx)=ddx(e2x((3A+2B)sin3x+(2A+3B)cos3x))

Apply the product rule (2),

y(x)={e2xddx[(3A+2B)sin3x+(2A+3B)cos3x]+[(3A+2B)sin3x+(2A+3B)cos3x]ddx(e2x)}={e2x[(3A+2B)cos3x(3)+(2A+3B)(sin3x)(3)]+[(3A+2B)sin3x+(2A+3B)cos3x](2e2x)}={e2x[(9A+6B)cos3x(6A+9B)sin3x]+e2x[(6A+4B)sin3x+(4A+6B)cos3x]}=e2x[(5A+12B)cos3x+(12A5B)sin3x]

Therefore, the second derivative of y is y=e2x[(5A+12B)cos3x+(12A5B)sin3x].

Substitute the values of y(x), y(x) and y(x) in y4y+13y,

y4y+13y={e2x[(5A+12B)cos3x+(12A5B)sin3x]4e2x[(3A+2B)sin3x+(2A+3B)cos3x]+13e2x(Acos3x+Bsin3x)}={e2x[(5A+12B)cos3x+(12A5B)sin3x]+e2x[(12A8B)sin3x+(8A12B)cos3x]+e2x(13Acos3x+13Bsin3x)}

Simplify further,

y4y+13y=e2x[(5A+12B8A12B+13A)cos3x+(12A5B+12A8B+13B)sin3x]=e2x[(0)cos3x+(0)sin3x]=e2x[0]=0

Hence, the function y=e2x(Acos3x+Bsin3x) satisfies the differential equation y4y+13y=0.

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