What is the solution for (dy)/(dx)+1/(x+4)y=x^{-2} ?

The solution for (dy)/(dx)+1/(x+4)y=x^{-2} is y=(ln(x))/(x+4)-4/(x(x+4))+(c_{1})/(x+4)

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(dy)/(dx)+1/(x+4)y=x^{-2}

\frac{d y}{d x} + \frac{1}{x + 4} y = x^{- 2}

y = \frac{ln ( x )}{x + 4} - \frac{4}{x ( x + 4 )} + \frac{c _{1}}{x + 4}

Solution steps

\frac{d y}{d x} + \frac{1}{x + 4} y = x^{- 2}

Solve linear ODE:

y = \frac{ln ( x )}{x + 4} - \frac{4}{x ( x + 4 )} + \frac{c _{1}}{x + 4}

y = \frac{ln ( x )}{x + 4} - \frac{4}{x ( x + 4 )} + \frac{c _{1}}{x + 4}

f (x) = \frac{9 x}{e ^{x}}

\int_{0}^{3} (2 x^{2} - 6 x) d x

x \to 0 lim (\frac{2}{x} - \frac{2}{∣ x ∣})

\int e^{- x} cos (5 x) d x

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

What is the solution for (dy)/(dx)+1/(x+4)y=x^{-2} ?
The solution for (dy)/(dx)+1/(x+4)y=x^{-2} is y=(ln(x))/(x+4)-4/(x(x+4))+(c_{1})/(x+4)