General Solutions to differential Equations
What are the steps to finding the general solution of a differential equation?
More specifically how do I solve:
$$\frac{dy}{dt} = t^{2}y + e^t$$
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$\begingroup$There is no simple answer to the first question. Finding the general solution in a closed form of a differential equation cannot always be done (consider $y'=\sin(t^2)$). Rather there are usually steps to finding general solutions of certain classes of differential equations.
This differential equation is a linear differential equation of the form
$\frac{dy}{dt}+p(t)y=q(t)$
where $p(t)=-t^2$ and $q(t)=e^t$.
To solve this class of differential equations we multiply both sides by what is called an integrating factor $\mu(t)=e^{\int p(t) dt}$.
Note then that $e^{\int p(t) dt}\frac{dy}{dt} +e^{\int p(t) dt}\frac{dy}{dt}p(t)y=\frac{d}{dt}\left(e^{\int p(t) dt}y\right)$. Thus
$$ e^{\int p(t) dt}y=\int q(t) dt. $$
One can solve from here for $y$. This method can be applied to the differential equation here since is a linear differential equation of the above form.
$\endgroup$ $\begingroup$Subtract $t^2y(t)$ from both sides :
$\frac{dy(t)}{dt} - t^2y(t) = e^t $
Let $ m(t) = e^{\int -t^2y(t)dt} = e^{-\frac{t^3}{3}} $
Then multiply both sides by $m(t)$ and substitute :
$ -e^{-\frac{t^3}{3}}t^2 = \frac{d}{dt}(e^{-\frac{t^3}{3}}) $
Apply the reverse product rule to the left hand side, then integrate and you will get :
$ \int \frac{d}{dt}(\frac{y(t)}{e^{\frac{t^3}{3}}})dt = \int e^{-\frac{t^3}{3}+t}dt $
Evaluating gives :
$\frac{y(t)}{e^{\frac{t^3}{3}}} = \int e^{-\frac{t^3}{3}+t}dt + c $
$c$ is an arbitrary constant of course. Divide both sides by the $m(t)$ we defined earlier and you will get :
$ y(t) = e^{\frac{t^3}{3}}(\int e^{-\frac{t^3}{3}+t}dt + c_1) $
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