Solving a simple system of ode

Question

Hi guys I was solving this fairly simple system, but for final answer I got something weird and just wanted to check if this was normal. $$ A=\begin{pmatrix} 0 & 0\\ 1 & t\\ \end{pmatrix} $$

and $\overline{x}^T = (x(t), y(t))$ then we want to solve $\dot{\overline{x}}=A\overline{x}$

what I did is treat them as two equations

\begin{align*} x'=0\\ y'=x+yt \end{align*}

From the first we got $x=c$ where $c$ is a constant then we can plug that in the second equation

$$y'=c+yt$$

then we can solve this using integration factor $u = e^{-1/2 t^2}$.

Thus we get

$$e^{-1/2 t^2}y' -te^{-1/2 t^2}y =c e^{-1/2 t^2}$$

Then we can integrate to obtain $e^{-1/2 t^2}y = \int c e^{-1/2 t^2}$. I think this looks weird because I cannot solve the integral and was hoping someone more experience take a look and tell me if this seems correct. Thank you!

Answer 1

It was already quite good. I am starting after you multiplied by $u$. $$\begin{align} e^{-1/2 t^2}y' -te^{-1/2 t^2}y &=c e^{-1/2 t^2} \\ \Longleftrightarrow e^{-t^2/2}y' +(e^{-t^2/2})'y &=c e^{-t^2/2} \\ \Longleftrightarrow (e^{-t^2/2}y)' &=c e^{-t^2/2} \\ \Longleftrightarrow e^{-t^2/2}y&=\int ce^{-t^2/2}=c\sqrt{\frac{\pi}{2}} \text{erf}\left( \frac{t}{\sqrt{2}} \right)+C \\ \Longleftrightarrow y&=e^{t^2/2} \left( c\sqrt{\frac{\pi}{2}} \text{erf}\left( \frac{t}{\sqrt{2}} \right)+C \right)\end{align}$$

• Second line: $-e^{-t^2/2}t=\frac{d}{dt} (-e^{-t^2/2})$
• Third line: Product rule
• Fourth line: Integrating both sides

Answer 2

We rearrange our differential equation $y'=c+yt$.

$$\frac{dy}{dt}-ty=c$$

Now it is in the general form of a first order linear non-homogeneous ODE:

$$\frac{dy}{dt}+P(t)y=Q(t)$$

Hence our integrating factor is:

$$\mu(t)=e^{\int P(t)~dt}$$

$$\mu(t)=e^{\int -t~dt}$$

Note that it is not neccessary to consider the constant of integration:

$$\mu(t)=e^{-\frac{t^2}{2}}$$

Hence, your integrating factor was not correct.

We now multiply both sides by our integrating factor $\mu(t)$.

$$e^{-\frac{t^2}{2}} \frac{dy}{dt}-t e^{-\frac{t^2}{2}}y =ce^{-\frac{t^2}{2}} $$

After substituting $-e^{-\frac{t^2}{2}} t=\frac{d}{dt} \left(e^{-\frac{t^2}{2}}\right)$and applying the reverse product rule, we obtain:

$$\int \frac{d}{dt} \left(e^{-\frac{t^2}{2}} y\right)~dt=\int {ce^{-\frac{t^2}{2}}}~dt$$

$$e^{-\frac{t^2}{2}} y=\int {ce^{-\frac{t^2}{2}}}~dt$$

We notice that $\int ce^{-\frac{t^2}{2}}~dt$ is not solvable in terms of elementary functions. We can either evaluate this by using the definition of the error function $\text{erf} (x)$ or by using Wolfram Alpha.

We evaluate this to:

$$\int c e^{-\frac{t^2}{2}}~dt=c \sqrt{\frac{\pi}{2}} \text{erf} \left(\frac{t}{\sqrt{2}}\right)+k$$

Where $k$ is the arbitrary constant of integration. Hence,

$$e^{-\frac{t^2}{2}} y=c \sqrt{\frac{\pi}{2}} \text{erf} \left(\frac{t}{\sqrt{2}}\right)+k$$

Therefore, our general solution is:

$$y(t)=e^{\frac{t^2}{2}}\left(c \sqrt{\frac{\pi}{2}} \text{erf} \left(\frac{t}{\sqrt{2}}\right)+k\right)$$

Please do not hesitate to ask if you have any doubts or questions.