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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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What have you tried? Indeed, this is a very simple ODE: it's a linear one. –  Romeo Dec 21 '12 at 20:10
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2 Answers

up vote 2 down vote accepted

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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You forgot the factor $e^{-k t^2/2}$. –  Robert Israel Dec 21 '12 at 20:20
    
@RobertIsrael Thanks. Fixed it now. –  user17762 Dec 21 '12 at 20:21
    
By the way, the integral is not elementary: it can be expressed using the error function. –  Robert Israel Dec 21 '12 at 20:28
    
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 $$\int_0^t e^{kx^2/2} \sin \left(\dfrac{\pi x}{10}\right) dx = \dfrac1{2i} \int_0^t \exp(kx^2/2 + i \pi x/10) dx - \dfrac1{2i} \int_0^t \exp(kx^2/2 - i \pi x/10) dx$$ –  user17762 Dec 21 '12 at 20:34
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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

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