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:
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?)
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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