Forced Harmonic Osscilators

Question

I have to compute the general solution and compute the solution given an initial value.

$ y'' + 3y' + 2y = t^2 $

$ y(0) = y'(0) = 0$

I understand the first thing I need to do is find the homogeneous equation solution.

$ y'' + 3y' + 2y = 0 $

Solving this gives me

$ y_h(t) = k_1e^{-2t} + k_2e^{-t} $

So my next step is finding the particular solution. So I need to find $ y_p(t) $.

I assume $y_p(t) = at^2$ since there is a $t^2$ on the right hand side of the initial equation.

So now I have:

$ y_p = at^2 $

$ y_p^{'} = 2at $

$ y_p^{''} = 2a $

So I plug this into the original equation and am given

$ 2a + 3(2at) + 2(at^2) = t^2 $$\rightarrow$$ 2a + 6a + 2at^2 = t^2$

Thus $ a = $$\frac{1}{2}$$ $ since $ 2at^2 = t^2 $

So now I go on to get the general solution: $ y(t) = k_1e^{-2t} + k_2e^{-t} + $$\frac{1}{2}$$t^2 $

and the derivative is $ y'(t) = -2k_1e^{-2t} + -k_2e^{-t} + t $

When I solve for $ k_1 $ and $ k_2 $ both are equal to zero (0). I am not sure whether this is right or not because then I am only left with $ y(t) = $$\frac{1}{2}$$t^2 $

I am just looking for someone to look over my work I am not sure whether this is right or not.

Answer 1

the particular solution should be $$y_p=At^2+Bt+C$$

Answer 2

You might notice that smaller powers appears when you take the derivative of $t^2$. So when you plug this in, your equation turned out to be

$$ 2a + 6at + 2at^2 = t^2 $$

You need this to be true for all $t$. This isn't possible, since even if $a = \frac12$, you're left with

$$ 1 + 3t = 0 $$

which obviously isn't true for all $t$.

The way to fix this is to add smaller powers of $t$ in your guess:

$$ y_p(t) = at^2 + bt + c $$

plugging this in gives

$$ (2a + 3b + 2c) + (6a + 2b)t + 2at^2 = t^2 $$

Now you can match the 3 coefficients to produce a system of equations

\begin{align} 2a &= 1 \\ 6a + 2b &= 0 \\ 2a + 3b + 2c &= 0 \end{align}