Help with ODE

Gold Member
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
$$2y'' + 3y' + y = t^2 + 85 sin(2t), y(0) = -2, y'(0) = 0.$$

My homogeneous solution is $$c1y1 + c2y2$$

$$y1 = e^\frac{-t}{2}, y2 = e^-^t, Wronskian = e^\frac{-3t}{2}+e^\frac{-t}{2}$$

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

1. Undetermined Coefficients
Assumed solution was $$At^2 + Bt + C + Dcos(2t) + Esin(2t)$$

$$yp' = 2At + B - 2Dsin(2t) + 2Ecos(2t)$$
$$yp'' = 2A - 4Dcos(2t) -4Esin(2t)$$

After pluggin this back into original equation $$2y'' + 3y' + y$$ I got the following equalities:
$$At^2 + Bt + C + 6At + 3B + 4A = t^2$$
$$cos(2t)(-7D + 6F) = 0$$
$$sin(2t)(-7F - 6D) = 85sin(2t)$$

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

$$A=\frac{3t}{3t+14}, B=\frac{-4t}{3t+14}$$

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

$$Yp = -y1 \int \frac{y2 g(t) dt}{Wronskian} + y2 \int \frac{y1 g(t) dt}{Wronskian}$$
where $$g(t) =\frac{ t^2 + 85sin(2t) } {2}$$
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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Homework Helper
Gold Member
In your undetermined coefficients solution, why do A and B depend on t? A and B are supposed to be constants, right?

The particular solution I get is:

$$y = t^2 - 6t + 14 -6\cos 2t -7 \sin 2t$$

Homework Helper
cronxeh said:
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
$$2y'' + 3y' + y = t^2 + 85 sin(2t), y(0) = -2, y'(0) = 0.$$

My homogeneous solution is $$c1y1 + c2y2$$

$$y1 = e^\frac{-t}{2}, y2 = e^-^t, Wronskian = e^\frac{-3t}{2}+e^\frac{-t}{2}$$

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

1. Undetermined Coefficients
Assumed solution was $$At^2 + Bt + C + Dcos(2t) + Esin(2t)$$

$$yp' = 2At + B - 2Dsin(2t) + 2Ecos(2t)$$
$$yp'' = 2A - 4Dcos(2t) -4Esin(2t)$$

I get the following equations:

4A+3B+C=0
6A+B=0
A=1
6E-7D=0
-7E-6D=85

Edit: Suppose I should compare this to the results reported by James . . .
Ok, I'm cool.

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Gold Member
James R said:
In your undetermined coefficients solution, why do A and B depend on t? A and B are supposed to be constants, right?

The particular solution I get is:

$$y = t^2 - 6t + 14 -6\cos 2t -7 \sin 2t$$

Doh.. of course! So since $$At^2 = t^2$$ then $$A=1$$ and I simply equate the other terms to 0. Thanks.