0
$\begingroup$

The question wants me to prove the following ODE via integrating factor method: $$ \frac{dx}{dt} + \frac{t}{(1+t^2)}x = \frac{1}{t(1+t^2)} $$

I got my integrating factor to be $ \mu(t)= \sqrt{1+t^2}$ and then it gets messy and my implicit solution is as follows:

$ \sqrt{1+t^2}x = \frac{1}{2}\ln{\frac{\sqrt{1+t^2}-1}{\sqrt{1+t^2}+1}} + C $

details of the workings are shown:

working

I'm not too sure whether I'm correct, but my peer told me that I could use some substitution $x=\tan{t} $ but I'm not sure why I need to do that substitution

And also if I am actually correct, the solution to that ODE could have a totally different expression if I were to use that tangent substitution right? (and that both expressions could be correct right? How do we verify if the expressions are equivalent?)

$\endgroup$
2
  • $\begingroup$ The integrating factor is correct. $\endgroup$ Sep 6, 2020 at 13:54
  • $\begingroup$ The solution is correct. Just to check the solution, replace it in the differential equation. $\endgroup$ Sep 6, 2020 at 13:59

1 Answer 1

1
$\begingroup$

If you set $t=\tan(s)$ and $y(s)=x(t)=x(\tan(s))$, then you get $$ y'(s)=x'(\tan(s))(1+\tan^2(s))=x'(t)(1+t^2)=-tx+\frac1t=-\tan(s)y(s)+\cot(s) $$ The integrating factor can now be determined as $\frac1{\cos(s)}$ and results in $$ \frac{y(s)}{\cos(s)}=\int\frac1{\sin(s)}ds=\int\frac{\sin s}{1-\cos^2(s)}ds \\ =\frac12\ln|1-\cos(s)|-\frac12\ln|1+\cos(s)|+C $$ After back-substitution this gives exactly your solution. If this way is simpler is most likely a rather subjective decision.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .