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

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

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x^3(dy)/(dx)+4x^2y=e^{-x}

x^{3} \frac{d y}{d x} + 4 x^{2} y = e^{- x}

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

Solution steps

x^{3} \frac{d y}{d x} + 4 x^{2} y = e^{- x}

Solve linear ODE:

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

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

\int 5 sin (3 x) sin (6 x) d x

\frac{15}{3 s + 3}

\int \frac{x ^{3} + 2 x}{x ^{4} + 4 x ^{2} + 3} d x

\frac{d y}{d t} + 0.7 t y = 7 t, y (0) = 7

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (2 x)

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