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

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

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(x^2-1)y^'+2y=(x+1)^2

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

y = \frac{x ( x + 1 )}{x - 1} + \frac{c _{1} ( x + 1 )}{x - 1}

Solution steps

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

Solve linear ODE:

y = \frac{x ( x + 1 )}{x - 1} + \frac{c _{1} ( x + 1 )}{x - 1}

y = \frac{x ( x + 1 )}{x - 1} + \frac{c _{1} ( x + 1 )}{x - 1}

\frac{d}{d p} (e^{p} (p + p p))

x \to - 4 lim (x^{2})

5

\int \frac{5}{t} d t

\int \frac{1}{3 ( x ^{2} + 1 )} d x

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