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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} - 4 y = sin (x)

y^{^{''}} + 4 y^{^{'}} + 32 y = 4 cos (4 t)

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 7

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

y^{^{''}} + 4 y = 3 sin (t), y (0) = 1, y^{^{'}} (0) = - 1

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