What is the solution for y^{'''}-y^{''}=xe^x ?

The solution for y^{'''}-y^{''}=xe^x is y=c_{1}e^x+c_{2}+c_{3}x+(e^xx^2-4e^xx+6e^x)/2

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

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

y = c_{1} e^{x} + c_{2} + c_{3} x + \frac{e ^{x} x ^{2} - 4 e ^{x} x + 6 e ^{x}}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + c_{3} x + \frac{e ^{x} x ^{2} - 4 e ^{x} x + 6 e ^{x}}{2}

y = c_{1} e^{x} + c_{2} + c_{3} x + \frac{e ^{x} x ^{2} - 4 e ^{x} x + 6 e ^{x}}{2}

(x^{2} + 4) y = 8, (- 2, 1)

\frac{( s + 2 )}{s ^{2} + s + 1}

\frac{d}{d x} (\frac{d y}{d x}) + 9 y = 0

\int e^{x} (x + 1) cos (x e^{x}) d x

y = (3 x - 4)^{5}

What is the solution for y^{'''}-y^{''}=xe^x ?
The solution for y^{'''}-y^{''}=xe^x is y=c_{1}e^x+c_{2}+c_{3}x+(e^xx^2-4e^xx+6e^x)/2