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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f (x) = (3 x - 2) (x + 4), (1, 5)

\frac{\partial}{\partial y} (x \cdot cos (y) \cdot sin (z))

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

\int \frac{( 3 x - 5 )}{x ^{2} + 4 x} d x

f (x) = 3 x^{2} - 7

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