What is the solution for (dy)/(dx)+1/(x+8)y=x^{-2} ?

The solution for (dy)/(dx)+1/(x+8)y=x^{-2} is y=(ln(x))/(x+8)-8/(x(x+8))+(c_{1})/(x+8)

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(dy)/(dx)+1/(x+8)y=x^{-2}

\frac{d y}{d x} + \frac{1}{x + 8} y = x^{- 2}

y = \frac{ln ( x )}{x + 8} - \frac{8}{x ( x + 8 )} + \frac{c _{1}}{x + 8}

Solution steps

\frac{d y}{d x} + \frac{1}{x + 8} y = x^{- 2}

Solve linear ODE:

y = \frac{ln ( x )}{x + 8} - \frac{8}{x ( x + 8 )} + \frac{c _{1}}{x + 8}

y = \frac{ln ( x )}{x + 8} - \frac{8}{x ( x + 8 )} + \frac{c _{1}}{x + 8}

\frac{x ^{2} - 2 x}{( x - 1 ) ^{2}}

f (x) = x^{4} + 9 e^{x}, (0, 9)

e^{x} y d y = (e^{- y} + e^{- 2 x - y}) d x

n = 0 \sum \infty \frac{n + 1}{n !}

\frac{d}{d θ} (3 + 3 sin (θ))

What is the solution for (dy)/(dx)+1/(x+8)y=x^{-2} ?
The solution for (dy)/(dx)+1/(x+8)y=x^{-2} is y=(ln(x))/(x+8)-8/(x(x+8))+(c_{1})/(x+8)