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I have the following equation:

$\frac{dx}{dt}+x=4\sin(t)$

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:

$4A\cos(t)-4B\sin(t)+4A\sin(t)+4B\cos(t)=4\sin(t)$

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: $x=f(h)+f(a)$

3 Answers 3

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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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    @Sean87 $N$o problem. Consider accepting this as the answer, if it answers your question!2012-05-20
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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

$4A\cos(t)-4B\sin(t)+4A\sin(t)+4B\cos(t)=4\sin(t)$

you are done. Collecting common factors you will get

$4(A+B)\cos(t)+4(A-B-1)\sin(t)=0$

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

$A-B=1$

$A+B=0$

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