Undetermined Coefficients Initial Value Question

[~Sam~]

Homework Statement

y''-4y=60e^4t

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 C₁e^0x+C₂e^4x where C1 and C2 are unknown coefficients to be found.

The particular solution is calculating by considering Y_p= AXe^4x
Differentiating that twice, and solving for Yp, I get Yp=15xe^4x

So the general solution for this is obtained by adding the homogeneous and the particular. So I get: C₁e^0x+C₂e^4x +15xe^4x

I differentiated this equation and got 4C₂+15e^4x+60x^4x

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)e^0x+-3.25e^4x+15xe^4x but this is wrong. Any ideas?

 

[dacruick]

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

 

[vela]

Your characteristic equation is wrong.

Homework Statement

y''-4y=60e^4t

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 C₁e^0x+C₂e^4x where C1 and C2 are unknown coefficients to be found.

The particular solution is calculating by considering Y_p= AXe^4x
Differentiating that twice, and solving for Yp, I get Yp=15xe^4x

So the general solution for this is obtained by adding the homogeneous and the particular. So I get: C₁e^0x+C₂e^4x +15xe^4x

I differentiated this equation and got 4C₂+15e^4x+60x^4x

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)e^0x+-3.25e^4x+15xe^4x 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)...