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} (a x^{2} e^{- x})

\frac{d}{d x} (ln (\frac{3}{x ^{2} + 9}))

f (x) = x^{x}, at x = 2

(x y^{2} - x) d x + (x^{2} y + y) d y = 0

\frac{\partial}{\partial x} (20 - 2 x + 1.5 - x + 3 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)