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In solving a problem involving differential equations, I come across the following:

$$\ddot{y} + 4\omega^2y = 2\omega gt\sin{\lambda} \equiv ct$$

The general solution is the general solution of the homogeneous equation and one particular solution of the inhomogeneous equation, i.e.,

$$y = \frac{c}{4\omega^2}t + A\sin{2\omega t} + B\cos{2\omega t}$$

I'm at a loss as to how it got to $y$. I can only think of the following:

$$\dot y + 4\omega^2yt = \omega gt^2\sin{\lambda} + C$$

which of course is nowhere near what I read. I'd appreciate if someone can point me in the right direction.

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I think it is fair to say that the particular solution $ct/(4\omega^2)$ is obvious, at least after one sees it. Differentiating twice will kill it, so $4\omega^2y=ct$ works. – André Nicolas Aug 24 '11 at 2:25
@Andre, I'm not sure if I get what you said. If you could explain it better, I'd appreciate it. Where is $\ddot y$ in $4\omega^2 y = ct$, as you mentioned? – Kit Aug 24 '11 at 2:31
I assume the way you got to the last display was by attempting to antidifferentiate both sides of the first display, but when you antidifferentiated $y$ and got $yt$, well, that would be correct if $y$ were a constant, but it isn't. This may not help point you in the right direction, but it may also be useful to you to see what's wrong with the direction you took. – Gerry Myerson Aug 24 '11 at 2:33
@Kit: Without any offense intended, just trying to give useful advice: If you believe that $\int y(t)dt = yt$, I think you need to brush up your calculus skills before you have a go at solving inhomogeneous second-order differential equations. That only works if $y$ is a constant independent of $t$. – joriki Aug 24 '11 at 2:47
Please do not (for this part) even think in terms of integrating. Let $y=\frac{c}{4\omega^2} t$. Does this $y$ satisfy the DE? Let's check by substituting. For this $y$, calculate $\ddot{y}$. What do you get? I get $0$. Calculate $4\omega^2y$. What do you get? I get $ct$. Calculate $\ddot{y}+4\omega^2y$. What do you get? I get $ct$. Let's now look at the right-hand side. It is $ct$. So the function $y$ we have been working with is a solution of the DE. – André Nicolas Aug 24 '11 at 2:54
up vote 2 down vote accepted

When they write $\omega gt \sin \lambda\equiv ct$ they mean to quickly introduce the substitution $c:=\omega g \sin\lambda$. Note furthermore that differentiating any polynomial twice will reduce its degree by two, so twice differentiating a linear map will send it to $0$ (i.e. annihilate, or kill, it). Hence if we introduce for our particular solution a linear map (as the RHS is just a linear map), we can simply choose $y=ct / (4\omega^2)$. This way


The homogeneous part to the solution will solve $\ddot{y}+4\omega^2y=0$. The characteristic equation has solutions $r=\pm 2\omega i$, so the general solution will be of the form

$$\frac{ct}{4\omega^2}+\alpha e^{2\omega i}+\beta e^{-2\omega i}$$

or, after a change of variables,

$$\frac{ct}{4\omega^2}+A\sin2\omega t+B\cos2\omega t.$$

Edit: Apparently your scratchwork indicates you believed one could simply integrate $y$ to get $yt+C$. I sternly agree with joriki's comment: this is false and it's going to be difficult to work with differential equations if you don't have the more elementary calculus fully understood yet.

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Thank you, @anon. I had a hunch that I needed to use some characteristic polynomial, but I didn't know how to articulate it in my question, or simply forgot about the technique, really :) – Kit Aug 24 '11 at 3:48

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