I just started studying differential equations, and I can't solve the following:
$$x'-2xt=2te^{2t}$$
I know it is a linear ordinary differential equation, and to solve it I have to do the change of variables $x=ce^{\int_{t_0}^{t}2s ds}$. I want to find the general solution, so I can't subsitute $t_0$ for any value like $0$ or $1$. If I was given the value of the function for a $t_0$, meaning: $x(t_0)=x_0$. I would know how to proceed. What can I do?
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Mutliply the equation by $e^{-t^2}$ $$x'-2xt=2te^{2t}$$ $$x'e^{-t^2}-2te^{-t^2}x=2te^{2t-t^2}$$ $$(xe^{-t^2})'=2te^{2t-t^2}$$ Integrate $$xe^{-t^2}=2 \int te^{2t-t^2}dt +K$$ $$x(t)=2e^{t^2} \int te^{2t-t^2}dt +Ke^{t^2}$$ You can evaluate the integral with the error function or keep that way $$x(t)= { \sqrt {\pi}}e^{t^2+1} \text {erf} (t-1)-e^{2t} +Ke^{t^2}$$ where $$\text {erf }(x)=\frac 2{\sqrt {\pi}}\int_0^x e^{-t^2} dt$$
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Using a standard method of finding an integrating factor one considers $$\frac{1}{\mu} \, \frac{d}{dt} \left( \mu \, y(t) \right) = y' + \left(\frac{d \, \ln \mu}{dt} \right) \, y.$$ Using this and comparing it to the differential equation in question one has $$\frac{d \, \ln \mu}{dt} = - 2 t = \frac{d(-t^2)}{dt},$$ or $\mu = e^{-t^2}$. Now, \begin{align} y' - 2 t \, y &= 2 t \, e^{2 t} \\ e^{t^2} \, D\left(e^{-t^2} \, y \right) &= 2t \, e^{2t} \\ D\left(e^{-t^2} \, y\right) &= 2t \, e^{2t - t^2} \end{align} and leads to $$y(t) = e^{t^2} \, \int^{t} 2u \, e^{- (u^2 - 2u)} \, du + c_{0}.$$
By completing the integration it can be shown that $$y(t) = c_{1} - e^{2 t} - \sqrt{\pi} \, e^{t^2 + 1} \, erf(1-t),$$ where $erf(x)$ is the error function.
Your ODE in the form of \begin{equation} x' + p(t)x = q(t) \end{equation} has the general solution is \begin{equation} x(t) = \frac{\int e^{\int p(t)}q(t) dt + C}{e^{\int p(t)}} \end{equation} where in your case, you have \begin{align} p(t) &= -2t \\ q(t) &= 2te^{2t} \end{align} The integration of $p(t)$ is \begin{equation} \int p(t) = -t^2 \end{equation} and hence \begin{equation} x(t) = \frac{2\int te^{-t^2+2t} dt + C}{e^{-t^2}} \tag{1} \end{equation} Notice that \begin{equation} \int te^{-t^2+2t} dt ={\displaystyle\int}\left(t-1\right)\mathrm{e}^{2t-t^2}\,\mathrm{d}t+{\displaystyle\int}\mathrm{e}^{2t-t^2}\,\mathrm{d}t \end{equation} The first integral is easy $\int f'e^{f} = e^{f}$and will give you \begin{equation} {\displaystyle\int}\left(t-1\right)\mathrm{e}^{2t-t^2}\,\mathrm{d}t =-\dfrac{\mathrm{e}^{2t-t^2}}{2} \end{equation} Now solving ${\displaystyle\int}\mathrm{e}^{2t-t^2}\,\mathrm{d}t$, you can complete the squares as such $${\displaystyle\int}\mathrm{e}^{2t-t^2}\,\mathrm{d}t={\displaystyle\int}\mathrm{e}^{1-\left(t-1\right)^2}\,\mathrm{d}t$$ Then do a change of variab;e $u = t-1$, to get \begin{equation} {\displaystyle\int}\mathrm{e}^{2t-t^2}\,\mathrm{d}t= =\dfrac{\sqrt{{\pi}}\mathrm{e}\operatorname{erf}\left(u\right)}{2} =\dfrac{\sqrt{{\pi}}\mathrm{e}\operatorname{erf}\left(t-1\right)}{2} \end{equation} Finally, we get \begin{equation} \int te^{-t^2+2t} dt =\dfrac{\sqrt{{\pi}}\mathrm{e}\operatorname{erf}\left(t-1\right)-\mathrm{e}^{2t-t^2}}{2}+C \end{equation}