What is the solution for y^{'''}-2y^'+4y=excos(x)+x^2 ?

The solution for y^{'''}-2y^'+4y=excos(x)+x^2 is

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y^{'''}-2y^'+4y=excos(x)+x^2

y^{^{'''}} - 2 y^{^{'}} + 4 y = e x cos (x) + x^{2}

y = c_{1} e^{- 2 x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{- 3 e x sin ( x ) + 4 e x cos ( x )}{25} - \frac{24 e}{125} sin (x) + \frac{7 e}{125} cos (x) + \frac{x ^{2} + x}{4} + \frac{1}{8}

Solution steps

y^{'''} (x) - 2 y^{'} (x) + 4 y = e x cos (x) + x^{2}

Solve linear ODE:

y = c_{1} e^{- 2 x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{- 3 e x sin ( x ) + 4 e x cos ( x )}{25} - \frac{24 e}{125} sin (x) + \frac{7 e}{125} cos (x) + \frac{x ^{2} + x}{4} + \frac{1}{8}

y = c_{1} e^{- 2 x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{- 3 e x sin ( x ) + 4 e x cos ( x )}{25} - \frac{24 e}{125} sin (x) + \frac{7 e}{125} cos (x) + \frac{x ^{2} + x}{4} + \frac{1}{8}

y^{^{''}} + 8 y^{^{'}} + 25 y = 150, y^{^{'}} (0) = 0, y (0) = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 t e^{t}, y (0) = 1, y^{^{'}} (0) = - 4

\frac{d ^{2} y}{d x ^{2}} + 14 \frac{d y}{d x} + 49 y = 4 e^{3 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 12 e x - 12

y^{^{''}} - y^{^{'}} - 2 y = 26 sin (x) + 40 x cos (x)

What is the solution for y^{'''}-2y^'+4y=excos(x)+x^2 ?
The solution for y^{'''}-2y^'+4y=excos(x)+x^2 is