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}}

\int csc^{2} (u) d u

n = 1 \sum \infty n! (0)^{n}

\frac{\partial}{\partial x} (y^{3} - 1)

\int 2 x \cdot sin (x^{2}) d 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)