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First Order Differential Equation : $$ \dfrac{df(t)}{dt}+ a f(t) = \dfrac{\sin(b t)}{\pi t} $$ $$ f(t)=? $$ I try to find solution in this way $$ \dfrac{dy}{dt}+ P(t) y = Q(t) $$ $$ y=e^{-\int P(t)dt} [\int Q(t) e^{\int P(t)dt} dt+c] $$ $$ y=e^{-at}[\int \frac{e^{at}sin(bt)}{t} dt+C ] $$ I can't Find solution for the integral. plz help me. thx so much

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Is this homework? If so you might want to tag it as such. –  user50407 Dec 21 '12 at 20:47
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Could you atleast show some efford in your questions? Show what you have tried instead of just giving us some homework assignment. –  sxd Dec 21 '12 at 20:48
    
At least the homogeneous solution should be easy to find.. –  malin Dec 21 '12 at 20:51
    
It's not homework.anyway thx so much. –  Amir Alizadeh Dec 21 '12 at 22:04
    
You can look at my last edit for some information about the integral. Were you expecting it to be elementary? –  user50407 Dec 21 '12 at 23:04

2 Answers 2

This is similar to the question you have asked here.

First note that $$\dfrac{df}{dt} + af = e^{-at} \dfrac{d}{dt} \left(e^{at} f\right) = \dfrac{\sin \left(bt\right)}{\pi t}$$ Hence, we have that $$\dfrac{d}{dt} \left(e^{at} f\right) = e^{at} \dfrac{\sin \left(bt\right)}{\pi t}$$ Hence, $$e^{at}f(t) = f(0) + \dfrac1{\pi} \int_0^t e^{ax} \dfrac{\sin \left(bx\right)}{x} dx$$ $$f(t) = e^{-at}f(0) + \dfrac{e^{-at}}{\pi} \int_0^t e^{ax} \dfrac{\sin \left(bx\right)}{x} dx$$

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thx so much, plz help me to find answer of the integral –  Amir Alizadeh Dec 21 '12 at 21:49

We note that this is a first order linear differential equation, so a good approach would be to use an integration factor:

We multiply the equation by an unknown $i(t)$ and want the LHS to equal $(i(t)f(t))'$.

$$i(t)f'(t)+ai(t) f(t)=i(t)\frac{\sin(bt)}{\pi t}$$

and we want $$(i(t)f(t))'=i(t)\frac{df}{dt}(t)+ai(t)f(t)$$

so by the product rule we get $i'(t)=ai(t)$, which is a separable differential equation which you can easily solve to obtain $i(t)$.

Now integrate both the LHS and RHS of the modified differential equation wrt $t$.

$$i(t)f(t)=\int i(t)\frac{\sin(bt)}{\pi t}\mathrm dt$$

And then the solution to the original differential equation is

$$f(t)=\dfrac{\int i(t)\frac{\sin(bt)}{\pi t}\mathrm d t}{i(t)}$$

This is a general way to solve this kind of problem but I think in this case you might run into some integrals which aren't expressable in terms of elementary functions.

EDIT I see you have edited in what you know and that you are only really having trouble with the final integral. Like I said, I don't believe it's expressable in terms of elementary functions. Here is a link to what Wolfram Alpha has to say about this: $\displaystyle \int \frac{e^{ax}\sin(bx)}{x}\mathrm dx$

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