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Is my solution to this first order ODE correct? (Integrating factor method)

Mathematics Asked by Ryan Seah on January 7, 2021

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?)

One Answer

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 $$ frac{y(s)}{cos(s)}=intfrac1{sin(s)}ds=intfrac{sin s}{1-cos^2(s)}ds \ =frac12ln|1-cos(s)|-frac12ln|1+cos(s)|+C $$ After back-substitution this gives exactly your solution. If this way is simpler is most likely a rather subjective decision.

Correct answer by Lutz Lehmann on January 7, 2021

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