I am confused about this example that I am trying to understand.

We solve the initial value problem $$x' = \frac{te^{x^2}}{x}, \space x(0) = 1, $$ and we do it "the usual way" by writing the equation as $x'xe^{-x^2} = t $ and integrating both sides. We end up with $x^2 = \mathrm{ln}\frac{1}{e^{-1}-t^2} \iff x = \pm \sqrt{\mathrm{ln}\frac{1}{e^{-1}-t^2}}.$ Then the text says that we know, because of the initial condition, which sign to choose in front of the root expression, so it should be $+\sqrt{\mathrm{ln}\frac{1}{e^{-1}-t^2}}.$ That is what confuses me.

I have learned that if there is a constant solution $x(t) \equiv k$, then any nonconstant solution cannot cross the line $x=k$. But from what I can gather, there is no constant solution to this equation. Because $x\equiv k \iff \frac{e^{x^2}}{x} = 0,$ which cannot happen for any $x$. Other than that I have no idea what the initial condition could possibly tell me about which sign to choose.


After integrating both sides, we find

$$-e^{-x^2}=t^2+C $$

but for $t=0$ we should have $x (0)=1$ thus $C=-e^{-1} $.

finally $$x^2 (t)=-\ln \left(-t^2+\frac {1}{e}\right) $$ $$=\ln \left(\frac {1}{-t^2+\frac {1}{e}}\right) $$

and $$x (t)=\sqrt {\ln \left( \frac {1}{-t^2+\frac {1}{e}} \right)}$$

if we take negative root, we will have $x (0)=-1$ instead of $1$.

  • $\begingroup$ It seems so obvious now that you did it. :p $\endgroup$
    – frej.mh
    Aug 14 '17 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.