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

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

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

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

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

Solution steps

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

Solve using substitution:

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

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

x \to 0 lim (1 - 1 - x)

2 x y^{^{'}} + y = 6 x

\frac{d y}{d x} = (y - 1) (1 - 2 y)

\int \frac{7 ln ( x )}{x ^{3}} d x

ln (2 + x)

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