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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (y^{2} - x^{2} - 10)

\frac{t d y}{d t} = t^{3} + 23 t^{3} y

\frac{( 6 x ^{2} - 5 ) ( x ^{2} - 6 )}{x ^{2} + 4}

y^{^{''}} - \frac{2 3}{3} y^{^{'}} + \frac{2}{9} = 0

\int_{0}^{2} (2 x - 1)^{2} d x

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