$$(1-t^2)x'-tx+t^2-1=0,\quad t\in(-1,1)$$

I have to determine the solution for this DE. I've tried dividing the equation with $(1-t^2)$, so it'd look like:


I've tried to first solve it as a homogeneous DE:




$x=e^{C_1}(1-t^2)^{-\frac 1{2}}=C(1-t^2)^{-\frac 1{2}}$

But then I'm stuck. I can't figure out what to use this to solve the non-homogeneous form. Any help?


You solved the homogeneous ODE but you also found that an integrating factor of your first order linear ODE is $(1-t^2)^{1/2}$.

Now multiply both sides of $x'(t)-\frac{t}{1-t^2}x(t)=1$ for such factor: $$(1-t^2)^{1/2}x'(t)-t(1-t^2)^{-1/2}x(t)=\frac{d}{dt}\left((1-t^2)^{1/2}x(t)\right)=(1-t^2)^{1/2}.$$ Hence, it remains to integrate $$(1-t^2)^{1/2}x(t)=\int (1-t^2)^{1/2}\, dt=\frac{1}{2}\left(t(1-t^2)^{1/2}+\arcsin(t)\right)+C$$ Finally $$x(t)=\frac{1}{2}\left(t+(1-t^2)^{-1/2}\arcsin(t)\right)+C(1-t^2)^{-1/2}.$$ Note that the term $C(1-t^2)^{-1/2}$ is the solution of the homogeneous ODE that you found in your work.

  • $\begingroup$ thank you so much! I figured out where I was getting stuck, it was the integrating factor part. So I'm guessing finding the solution for hom-ODE necessary wasn't all that necessary, was it? $\endgroup$ – bird Jul 9 at 8:49
  • $\begingroup$ @bird Yes, once you that have the integrating factor you can find the complete solution. $\endgroup$ – Robert Z Jul 9 at 8:52
  • $\begingroup$ You have to integrate $-t/(1-t^2)$ i.e. the coefficient of $x$ in $x'(t)-\frac{t}{1-t^2}x(t)=1$. $\endgroup$ – Robert Z Jul 9 at 9:00
  • $\begingroup$ again thanks for the help!! found the error in my calc! $\endgroup$ – bird Jul 9 at 9:02

Try the following ansatz: $$ x = x_{hom} f $$ where $x_{hom}$ is the solution to the homogenious equation you found. Fom this you get: $$ (x)^\prime-\frac{t}{1-t^2}x = f^\prime x_{hom} + f\left(x^\prime_{hom}-\frac{t}{1-t^2}x_{hom}\right) = f^\prime x_{hom} = 1 $$ since you know $x_{hom}$ you may now find $f$, and thus $x$, by integration.


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