- #1

cronxeh

Gold Member

- 1,004

- 10

Ok I'm stuck on this problem - and it didnt particularly look hard when I started doing it, oh say about a good 2 hours ago.. hmm

Solve the IVP

[tex]2y'' + 3y' + y = t^2 + 85 sin(2t), y(0) = -2, y'(0) = 0.[/tex]

My homogeneous solution is [tex]c1y1 + c2y2[/tex]

[tex]y1 = e^\frac{-t}{2}, y2 = e^-^t, Wronskian = e^\frac{-3t}{2}+e^\frac{-t}{2} [/tex]

So for a particular solution I've tried 2 different methods

1. Undetermined Coefficients

Assumed solution was [tex]At^2 + Bt + C + Dcos(2t) + Esin(2t)[/tex]

[tex]yp' = 2At + B - 2Dsin(2t) + 2Ecos(2t)[/tex]

[tex]yp'' = 2A - 4Dcos(2t) -4Esin(2t)[/tex]

After pluggin this back into original equation [tex]2y'' + 3y' + y[/tex] I got the following equalities:

[tex]At^2 + Bt + C + 6At + 3B + 4A = t^2[/tex]

[tex]cos(2t)(-7D + 6F) = 0[/tex]

[tex]sin(2t)(-7F - 6D) = 85sin(2t)[/tex]

Fiddling around with the equalities I got C=0, D=-6, F=-7

[tex]A=\frac{3t}{3t+14}, B=\frac{-4t}{3t+14}[/tex]

After assembling the whole thing together I don't get the right hand side after integrating to check the solutions, and so even when I derive the y=yp+yh solution to get my constants for homogeneous equation I get wrong coefficients c1 and c2 for yh - an infinite number of them, so I'm pretty sure my solution is wrong

Should I have multiplied the assumed solution by t anywhere?

2. Variation of Parameters

[tex]Yp = -y1 \int \frac{y2 g(t) dt}{Wronskian} + y2 \int \frac{y1 g(t) dt}{Wronskian}[/tex]

where [tex]g(t) =\frac{ t^2 + 85sin(2t) } {2}[/tex]

The result was extremely long and contained a lot of exponentials, and I assumed it to be incorrect, although any comment on the

Solve the IVP

[tex]2y'' + 3y' + y = t^2 + 85 sin(2t), y(0) = -2, y'(0) = 0.[/tex]

My homogeneous solution is [tex]c1y1 + c2y2[/tex]

[tex]y1 = e^\frac{-t}{2}, y2 = e^-^t, Wronskian = e^\frac{-3t}{2}+e^\frac{-t}{2} [/tex]

So for a particular solution I've tried 2 different methods

1. Undetermined Coefficients

Assumed solution was [tex]At^2 + Bt + C + Dcos(2t) + Esin(2t)[/tex]

[tex]yp' = 2At + B - 2Dsin(2t) + 2Ecos(2t)[/tex]

[tex]yp'' = 2A - 4Dcos(2t) -4Esin(2t)[/tex]

After pluggin this back into original equation [tex]2y'' + 3y' + y[/tex] I got the following equalities:

[tex]At^2 + Bt + C + 6At + 3B + 4A = t^2[/tex]

[tex]cos(2t)(-7D + 6F) = 0[/tex]

[tex]sin(2t)(-7F - 6D) = 85sin(2t)[/tex]

Fiddling around with the equalities I got C=0, D=-6, F=-7

[tex]A=\frac{3t}{3t+14}, B=\frac{-4t}{3t+14}[/tex]

After assembling the whole thing together I don't get the right hand side after integrating to check the solutions, and so even when I derive the y=yp+yh solution to get my constants for homogeneous equation I get wrong coefficients c1 and c2 for yh - an infinite number of them, so I'm pretty sure my solution is wrong

Should I have multiplied the assumed solution by t anywhere?

2. Variation of Parameters

[tex]Yp = -y1 \int \frac{y2 g(t) dt}{Wronskian} + y2 \int \frac{y1 g(t) dt}{Wronskian}[/tex]

where [tex]g(t) =\frac{ t^2 + 85sin(2t) } {2}[/tex]

The result was extremely long and contained a lot of exponentials, and I assumed it to be incorrect, although any comment on the

**correct**answer you got would be appreciated
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