What is the solution for xy^'+4y=((2e^{-4x}))/(x^3) ?

The solution for xy^'+4y=((2e^{-4x}))/(x^3) is y=-(e^{-4x})/(2x^4)+(c_{1})/(x^4)

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xy^'+4y=((2e^{-4x}))/(x^3)

x y^{^{'}} + 4 y = \frac{( 2 e ^{- 4 x} )}{x ^{3}}

y = - \frac{e ^{- 4 x}}{2 x ^{4}} + \frac{c _{1}}{x ^{4}}

Solution steps

x y^{'} (x) + 4 y = \frac{( 2 e ^{- 4 x} )}{x ^{3}}

Solve linear ODE:

y = - \frac{e ^{- 4 x}}{2 x ^{4}} + \frac{c _{1}}{x ^{4}}

y = - \frac{e ^{- 4 x}}{2 x ^{4}} + \frac{c _{1}}{x ^{4}}

\frac{d}{d x} (\frac{x + 2}{x - 1})

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

n = 0 \sum \infty 3.2 5^{n}

f (x) = \frac{1}{3 + 2 x}, at x = 1

What is the solution for xy^'+4y=((2e^{-4x}))/(x^3) ?
The solution for xy^'+4y=((2e^{-4x}))/(x^3) is y=-(e^{-4x})/(2x^4)+(c_{1})/(x^4)