1
$\begingroup$

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.

  • 0
    @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

1 Answers 1

2

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.

  • 0
    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