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}}

y = - 3 x^{2} + 10 x + 2, at x = 1

\frac{t d u}{d t} = t^{2} + 3 u, t > 0, H (6) = 180

5 - 2 x^{2}

\frac{d y}{d x} = 0.05 x - 450

\frac{0.6}{s + 10}

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)