What is the solution for (1-t^2)y^{''}-2ty^'+2y=0 ?

The solution for (1-t^2)y^{''}-2ty^'+2y=0 is y=c_{1}-1/2 c_{1}tln(t+1)+1/2 c_{1}tln(t-1)+c_{2}t

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(1-t^2)y^{''}-2ty^'+2y=0

(1 - t^{2}) y^{^{''}} - 2 t y^{^{'}} + 2 y = 0

y = c_{1} - \frac{1}{2} c_{1} t ln (t + 1) + \frac{1}{2} c_{1} t ln (t - 1) + c_{2} t

Solution steps

(1 - t^{2}) y^{''} (t) - 2 t y^{'} (t) + 2 y = 0

Solve using substitution:

y = c_{1} - \frac{1}{2} c_{1} t ln (t + 1) + \frac{1}{2} c_{1} t ln (t - 1) + c_{2} t

y = c_{1} - \frac{1}{2} c_{1} t ln (t + 1) + \frac{1}{2} c_{1} t ln (t - 1) + c_{2} t

What is the solution for (1-t^2)y^{''}-2ty^'+2y=0 ?
The solution for (1-t^2)y^{''}-2ty^'+2y=0 is y=c_{1}-1/2 c_{1}tln(t+1)+1/2 c_{1}tln(t-1)+c_{2}t