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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.2011-08-24
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    @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?2011-08-24
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    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.2011-08-24
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    @Gerry, it's quite clear that $y$ is a function of $t$, so $\int y(t)dt = yt$, but it isn't anywhere near the given solution. So, how should I proceed?2011-08-24
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    @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 $t$.2011-08-24
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    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.2011-08-24
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    @joriki, thank you, I take your advice well :) I was also rather shaky with $\int y(t)dt = yt$. It was my hope with this question that any offered solution might help me recover that missing link in my calculus skills. Actually I grazed over the quoted statements and was able to follow the consequent parts of the solution. So, if you don't mind helping with what you have, don't worry, I think I can grok it somehow. It's been ~3 years since undergrad :)2011-08-24
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    @Kit: anon has already provided a solution. You can also check out [the Wikipedia article](http://en.wikipedia.org/wiki/Linear_differential_equation#Nonhomogeneous_equation_with_constant_coefficients) to understand the theory, though the very general treatment there might be a bit of an overkill for this case; [here](http://www.efunda.com/math/ode/linearode_consthomo.cfm)'s a more specific treatment of the second-order homogeneous case. The general solution of the inhomogeneous equation is the sum of any solution of the inhomogeneous equation and all solutions of the homogeneous equation.2011-08-24

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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

$$(0)+4\omega^2\left(\frac{ct}{4\omega^2}\right)=ct.$$

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 :)2011-08-24