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}

f (x) = (2 x + 3)^{2}

\int_{0}^{2} 2 x^{3} x^{2} + 4 d x

x \to 1 lim (x^{3} - 3)

\int arcsin (8 x) d x

f (x) = x^{2} + x, at x = - 1, x = 2

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)