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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

sin (x) \frac{d y}{d x} + y cos (x) = 0, y (\frac{π}{2}) = 1

\frac{d y}{d x} = x^{3} 6 y

\frac{d S}{d t} = 0.05 - \frac{S}{( \frac{945}{2} )}

\frac{d x}{d t} = 4 x t^{2}

\frac{d y}{d t} = k (a - y)

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