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

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

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

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

y = - \frac{e ^{- 4 x}}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

Solution steps

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

Solve linear ODE:

y = - \frac{e ^{- 4 x}}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

y = - \frac{e ^{- 4 x}}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

\frac{d}{d x} (y (1 + e^{x}))

x \to 1 lim (0.9 9^{2 - 1})

\frac{d}{d x} (- \frac{13}{x ^{5}})

\frac{\partial}{\partial y} (- sin (x))

3 x arcsin (x)

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