Is there an analytical solution to this differential equation?
Emily Wong 
I am not very knowledgeable in the field so I apologize in advance if this question might look naive. But is there an analytical solution to a differential equation that looks like this :
$$\frac {\mathrm{d}y}{\mathrm{d}t} = a\cdot v(t) - \mathrm{b}\cdot y $$
My issue here is the function $v(t)$ and how I should deal with it.
Thanks
$\endgroup$ 43 Answers
$\begingroup$The solution of the homogeneous equation
$$y'(t)=-by(t)$$ is easily found to be $y(t)=Ce^{-bt}$.
Then by variation of the constant, $y(t)=c(t)e^{-bt}$ yields
$$c'(t)e^{-bt}-bc(t)e^{-bt}=av(t)-bc(t)e^{-bt}$$
or
$$c'(t)=av(t)e^{bt},$$
$$c(t)=\int av(t)e^{bt}dt+C,$$
$$y(t)=\left(\int av(t)e^{bt}dt+C\right)e^{-bt}.$$
Obviously, the equation has an analytical solution when the integral has one.
If the unknown is the function $v$, then
$$v(t)=\frac{y'(t)+by(t)}a.$$
$\endgroup$ 1 $\begingroup$Yes,there is:
$$y(t)=ce^{-bt}+ae^{-bt}\int_1^t e^{bx} \nu(x)\,dx$$
where $c$ is a constant.
$\endgroup$ $\begingroup$If I understand correctly, your equation is of the from $$ \frac{\mathrm{d}y}{\mathrm{d}y} = -by(x) + h(x) $$
A solution then is available (, just setting $g(x) = 1, f_1(x) = -b , f_0(x)=h(x) $)
$$ y = Ce^{-bx} + e^{-bx} \int e^{-bx} h(x)\mathrm{d}x $$
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