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How to solve the following first order differential equation?

$\dfrac{dy}{dt}+kty(t) = \dfrac{\sin(\pi t/10)}{\pi}$

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First note that $\dfrac{dy}{dt} + kty = e^{-kt^2/2} \left(e^{kt^2/2} \dfrac{dy}{dt} + e^{kt^2/2}kty \right) = e^{-kt^2/2} \dfrac{d}{dt}\left(e^{kt^2/2} y\right) = \dfrac{\sin \left(\dfrac{\pi t}{10}\right)}{10}$ Hence, we have that $\dfrac{d}{dt}\left(e^{kt^2/2} y\right) = e^{kt^2/2} \dfrac{\sin \left(\dfrac{\pi t}{10}\right)}{10}$ Hence, $e^{kt^2/2}y(t) = y(0) + \dfrac1{10} \int_0^t e^{kx^2/2} \sin \left(\dfrac{\pi x}{10}\right) dx$ $y(t) = y(0)e^{-kt^2/2} + \dfrac{e^{-kt^2/2}}{10} \int_0^t e^{kx^2/2} \sin \left(\dfrac{\pi x}{10}\right) dx$

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    Adding to @RobertIsrael's comment, the integral $$\int_0^t e^{kx^2/2} \sin \left(\dfrac{\pi x}{10}\right) dx$$ can be written in terms of error function, by replacing $\sin(\pi x/10)$, as $\dfrac{\exp(i \pi x/10) - \exp(-i \pi x/10)}{2i}$ and hence $\int_0^t e^{kx^2/2} \sin \left(\dfrac{\pi x}{10}\right) dx = \dfrac1{2i} \int_0^t \exp(kx^2/2 + i \pi x/10) dx - \dfrac1{2i} \int_0^t \exp(kx^2/2 - i \pi x/10) dx$2012-12-21
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This is a first order linear differential equation. You can solve it by using the integrating factor method. Here is a reference: http://www.sosmath.com/diffeq/first/lineareq/lineareq.html