What is the solution for y^{'''}-y^{''}-y^'+y=6e^x ?

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

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y^{'''}-y^{''}-y^'+y=6e^x

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 6 e^{x}

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

Solution steps

y^{'''} (x) - y^{''} (x) - y^{'} (x) + y = 6 e^{x}

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 4 - e^{x} + e^{5 x}

y^{^{''}} + 4 y = 3, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 4 = e^{- t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 3

(D^{2} + 4) y = sinh (x) sin (2 x)

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