What is the solution for xy^{'''}-xy^'=x^4 ?

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

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xy^{'''}-xy^'=x^4

x y^{^{'''}} - x y^{^{'}} = x^{4}

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - \frac{x ^{4}}{4} - 3 x^{2}

Solution steps

x y^{'''} (x) - x y^{'} (x) = x^{4}

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - \frac{x ^{4}}{4} - 3 x^{2}

y = c_{1} + c_{2} e^{- x} + c_{3} e^{x} - \frac{x ^{4}}{4} - 3 x^{2}

y^{^{''}} - y^{^{'}} = 5 e^{x}

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 32 y = (x + 1) e^{x}

2 y^{^{''}} + y^{^{'}} - y = e^{3 t}, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} = y + e^{2 x} \cdot e^{e^{x}}

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