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

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

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(x^2-1)((dy)/(dx))+2y=(x+1)^2

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

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

Solution steps

(x^{2} - 1) (\frac{d y}{d 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{\partial}{\partial x} (\frac{y}{x} ln (x))

n = 1 \sum \infty (cos (2.6))^{n}

\int (5 x + 1)^{\frac{1}{2}} d x

y^{^{''}} + 4 y = 3 cos (2 t)

\frac{d}{d n} (n^{n})

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