Finding a second linearly independent solution to a differential equation

There's a method for doing this, called reduction of order. I have linked to the Wikipedia article, which should get you started.

First, your equation I suppose in the shape of: $$x^{''}-{{2t}\over{1−t^2}}x^{'} +{{2}\over{1−t^2}}x=0$$ Your first solution is $x(t) = t$. So $x^{'}(t) = 1$ and $x^{''}(t) = 0$. Substituting into left hand part of the equation we get $0-{{2t}\over{1−t^2}}\cdot{1} +{{2}\over{1−t^2}}t = {{-2t+2t}\over{1-t^2}} = 0$ and this equals to right hand part of the equation, this means also {0}. So solution x(t) = t is one solution of the equation.
The second solution of the equation we assume in the shape of $x(t) = e^{kt}$, where k is value independent on t. So $x^{'}(t) = k{e^{kt}}$ and $x^{''}(t) = k^2{e^{kt}}$. Substituting into the equation we get: $k^2{e^{kt}}-{{2t}\over{1−t^2}}k{e^{kt}} +{{2}\over{1−t^2}}e^{kt}=0$. We need to identify {k} value. Because e^{kt} is not $0$ we can modify the form of the equation to $k^2-{{2t}\over{1−t^2}}k +{{2}\over{1−t^2}}=0$ and we now have for {k} the quadratic equation. Solving this quadratic equation we get $k_{1,2} = {{t\pm\sqrt{3t^2-2}}\over{1-t^2}}$.
So now we have two more terms which satisfy the equation: $$x(t)= {{t\pm\sqrt{3t^2-2}}\over{1-t^2}}$$(one with the $+$ sign in exponent a second with $-$ sign.
Whether this solution meets with everyone again by inserting:
${(t\pm\sqrt{3t^2-2})^2\over{{1-t^2}^2}}-{{2t}\over{1-t^2}}{{t\pm\sqrt{3t^2-2}}\over{1-t^2}}+{{2}\over{1-t^2}} = {{t^2\pm{2t\sqrt{3t^-2}}+3t^2-2-2t^2\mp{2t\sqrt{3t^-2}}+2-2t^2}\over{1-t^2}}={{0}\over{1-t^2}} = 0$. Is also equals to right hand part of equation. So both these solution also complies with the equation.
Including your solution we have three possible $x$: $$x(t) = t$$ $$x(t)= {{t+\sqrt{3t^2-2}}\over{1-t^2}}$$ $$x(t)= {{t-\sqrt{3t^2-2}}\over{1-t^2}}$$