# determine solution for differential equation

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

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

$$x'-\frac{t}{1-t^2}x=1$$

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

$$\frac{1}{x}dx-\frac{t}{1-t^2}dt=0$$

$$\ln|x|+\ln(1-t^2)^{\frac{1}{2}}=C_1$$

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

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

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.
• You have to integrate $-t/(1-t^2)$ i.e. the coefficient of $x$ in $x'(t)-\frac{t}{1-t^2}x(t)=1$. – Robert Z Jul 9 at 9:00
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.