# Solve y'' - y' = t - e^{-t} using variation of parameters

#### Eigendorf

##### New member
Hello, I'm trying to solve a differential equation using variation of parameters, but I'm not sure what I am doing wrong. The differential equation can also be solved by the method of undetermined coefficients, but the homework wants us to practice variation of parameters.

Any how the question:

Solve y'' - y' = t - e^-t

using variation of parameters

I've solved this using the method of undetermined coefficients and the answer i get is

C1 + C2*e^t - (t^2)/2 - T - (e^-t)/2

when I solve using the variation of parameters the answer I get is similar, can you please let me know what I'm doing wrong.

First I solved for Yh and I got Yh = C1 + C2*e^t

Then I solved for the Wronskian

Wronskian = det [1 e^t]
[0 e^t]

Wronskian = e^t

Then I solved W1 and W2

W1 = [0 e^t]
[t - e^-t e^t]

W1 = 1 - t*e^-t

W2 = [1 0]
[0 t - e^-t]

W2 = t - e^-t

Solve u1 and u2

u1 = integral [1 - t*e^-t]/(e^t) dt

u1 = -e^-t + te^(-2t)/2 + e^(-2*t)/4

u2 = integral [(t- e^-t)/e^t] dt

u2 = e^(-2t)/2 - te^-t - e^-t

solve Yp = u1y1 + u2y2

Yp = -e^-t/2 + te^(-2t)/2 + e^(-2*t)/4 -t - 1

combine Yh and Yp

Y(t) = C1 + C2e^t - e^-t/2 + te^(-2t)/2 + e^(-2t)/4 - t - 1

This is close to the answer i got from method of undetermined coefficients which was

Y(t) = C1 + C2e^t - e^-t/2 + te^(-2t)/2 + e^(-2t)/4 - (t^2)/2 - t

I seem to be off on the last two terms.. Any help would be appreciated on this question.

Thank you!

edit.. I see where I went wrong, I flipped a negative sign

W1 should be 1 - t*e^t

instead of W1 = 1 - t*e^-t

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