A few of my classmates and I are trying to solve the following differential equation: $x'' - x'\sin(t) - x\cos(t) = 0$ using the variation of parameters. However, we're at a bit of a loss because we struggled to find a fundamental solution, and at this point we are not even sure if this is the correct method to use. We tried using $x(t) = A\cos(t) + B\sin(t)$, $e^{at}\cos(t)$, and $e^{at}\sin(t)$ and substituting $x, x'$, and $x''$ from each back into the original equation, but none of these are working. Please provide some direction as to if this is even the correct way to approach the problem. Kind Regards, GingerKittyLover
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Note that $$0=x''(t) - x'(t)\sin(t) - x(t)\cos(t)=D(x'(t)-x(t)\sin(t))$$ and it follows that $$x'(t)-x(t)\sin(t)=c$$ which is a linear ODE of the first order.
Can you take it from here?
Robert Z
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Thank you very much! Yes, this helps a lot (it's a much simpler way to represent the equation) and we can take it from here. :) – CinnamonKittyLover Mar 15 '19 at 23:01