What is the solution for y^'+4/(x^2-1)y=0 ?

The solution for y^'+4/(x^2-1)y=0 is y=(e^{c_{1}}x^2+2e^{c_{1}}x+e^{c_{1}})/(x^2-2x+1)

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y^'+4/(x^2-1)y=0

y^{^{'}} + \frac{4}{x ^{2} - 1} y = 0

y = \frac{e ^{c_{1}} x ^{2} + 2 e ^{c_{1}} x + e ^{c_{1}}}{x ^{2} - 2 x + 1}

Solution steps

y^{'} (x) + \frac{4}{x ^{2} - 1} y = 0

Solve separable ODE:

y = \frac{e ^{c_{1}} x ^{2} + 2 e ^{c_{1}} x + e ^{c_{1}}}{x ^{2} - 2 x + 1}

y = \frac{e ^{c_{1}} x ^{2} + 2 e ^{c_{1}} x + e ^{c_{1}}}{x ^{2} - 2 x + 1}

\int 4 t^{2} d t

x \to 1 lim (- \frac{1}{x - 1})

\frac{7 + t + t ^{2}}{t}

y = \frac{w ^{4} + 5 w ^{2} + w}{w ^{2}}

y = \frac{2}{x}, y = 8 x, y = \frac{1}{2} x

What is the solution for y^'+4/(x^2-1)y=0 ?
The solution for y^'+4/(x^2-1)y=0 is y=(e^{c_{1}}x^2+2e^{c_{1}}x+e^{c_{1}})/(x^2-2x+1)