# Forced Harmonic Osscilators

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.

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

• Why? I thought we only needed t^2. – Andy Cohen Apr 25 at 21:57
• This according to the undetermined coefficient method – E.H.E Apr 25 at 21:58
• There is a chart at Paul's Online Notes that gives the form of the initial guess for $y_p$. – John Wayland Bales Apr 25 at 22:01

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}