# Undetermined Coefficients Initial Value Question

~Sam~

## Homework Statement

y''-4y=60e4t

y(0)=9 and y'(0)=2

## Homework Equations

none in particular

## The Attempt at a Solution

First I solved for the auxiliary equation.

r2-4r
r(r-4)
r=0, 4

So the general solution for the homogenous form is C1e0x+C2e4x where C1 and C2 are unknown coefficients to be found.

The particular solution is calculating by considering Yp= AXe4x
Differentiating that twice, and solving for Yp, I get Yp=15xe4x

So the general solution for this is obtained by adding the homogeneous and the particular. So I get: C1e0x+C2e4x +15xe4x

I differentiated this equation and got 4C2+15e4x+60x4x

So then I got 4C2+15=2, so C2=-3.25
Also, from the general solution C1+C2=9, so C1=12.25

The final answer I got was (12.25)e0x+-3.25e4x+15xe4x but this is wrong. Any ideas?

## Answers and Replies

dacruick
you are not using the method of undeteremined coefficients i don't think. try searching that up on google.

Staff Emeritus
Homework Helper
Your characteristic equation is wrong.

## Homework Statement

y''-4y=60e4t

y(0)=9 and y'(0)=2

## Homework Equations

none in particular

## The Attempt at a Solution

First I solved for the auxiliary equation.

r2-4r
r(r-4)
r=0, 4

So the general solution for the homogenous form is C1e0x+C2e4x where C1 and C2 are unknown coefficients to be found.

The particular solution is calculating by considering Yp= AXe4x
Differentiating that twice, and solving for Yp, I get Yp=15xe4x

So the general solution for this is obtained by adding the homogeneous and the particular. So I get: C1e0x+C2e4x +15xe4x

I differentiated this equation and got 4C2+15e4x+60x4x

So then I got 4C2+15=2, so C2=-3.25
Also, from the general solution C1+C2=9, so C1=12.25

The final answer I got was (12.25)e0x+-3.25e4x+15xe4x but this is wrong. Any ideas?

~Sam~
Your characteristic equation is wrong.

OHHH I see...thanks for pointing that out.. so the equation would be (r+2)(r-2)....