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}}

d er i v a t i v e y = \frac{1}{( 4 x ^{2} + 3 ) ^{5}}

\int (2 x^{3} + 5 x^{2} - 2 x + 7) d x

\frac{\partial}{\partial y} (x y + ln (\frac{x}{y}))

\int x^{4} d x

\int - 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)