What is the solution for (1-x^2)y^{''}-2xy^'+2y=0 ?

The solution for (1-x^2)y^{''}-2xy^'+2y=0 is y=c_{1}-1/2 c_{1}xln(x+1)+1/2 c_{1}xln(x-1)+c_{2}x

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(1-x^2)y^{''}-2xy^'+2y=0

(1 - x^{2}) y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

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

Solution steps

(1 - x^{2}) y^{''} (x) - 2 x y^{'} (x) + 2 y = 0

Solve using substitution:

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

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

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\int sin (x) sec^{2} (x) d x

What is the solution for (1-x^2)y^{''}-2xy^'+2y=0 ?
The solution for (1-x^2)y^{''}-2xy^'+2y=0 is y=c_{1}-1/2 c_{1}xln(x+1)+1/2 c_{1}xln(x-1)+c_{2}x