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I have the following equation:


For solving, I find the homogenous part as: $$f(h)=C*e^{-t}$$

Then finding $f(a)$ and $df(a)$: $$f(a)=4A\sin(t)+4B\cos(t)$$ $$df(a)=4A\cos(t)-4B\sin(t)$$

Substituting in orginal equation:


I have to find numerical values of $A$ and $B$ but I absoloutly have no idea how can I solve this, I am also not sure if the steps I did are correct or not. Would somone please help with this equation?

The final answer should be substituted in: $$x=f(h)+f(a)$$

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up vote 4 down vote accepted

Another approach would be to consider the given equation as one of the form

$$ \frac{dx}{dt}+Px=Q $$

where $P$ and $Q$ are functions of $t$. Such equation can be solved by multiplying both sides by the integrating factor $e^{\int P\,dt}$.

Applying this to your equation, we get $$ \begin{align} e^{t}\frac{dx}{dt}+e^t x=e^t 4 \sin{t}\\ \frac{d}{dt}\left(xe^t\right)=4\,e^t \sin{t}\\ xe^t=4\int e^t \sin{t}\ dx\\ \end{align} $$

Solving the above by integration by parts we get

$$ \begin{align} xe^t&=\frac{4e^t(\sin t -\cos t)}{2} + c\\ x&=2\sin t -2\cos t+ce^{-t} \end{align} $$

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Thanks, this actually makes it easier – Sean87 May 20 '12 at 17:09
@Sean87 No problem. Consider accepting this as the answer, if it answers your question! – Milosz Wielondek May 20 '12 at 18:15

multiples of $\cos(t)$ can't add to make $\sin(t)$, and vice versa. So this can be split into two equations: $$4A\cos(t)+4B\cos(t)=0$$ $$4A\sin(t)-4B\sin(t)=4\sin(t)$$ simplifying we get $$A+B=0$$ $$A-B=1$$ and so $$A=\frac{1}{2}, B=-\frac{1}{2}$$

So your final solution is $$x=f(h)+f(a)=Ce^{-t}+2\sin(t)-2\cos(t)$$

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At the stage


you are done. Collecting common factors you will get


but this must be independent and so you have a system of equations



giving $A=-B=\frac{1}{2}$.

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