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

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

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

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

y = c_{1} e^{- 2 x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{3 e ^{x} x sin ( x ) - e ^{x} x cos ( x )}{20} + \frac{x ^{2} + x}{4} + \frac{1}{8} + \frac{1}{40} sin (2 x) + \frac{3}{40} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{3 e ^{x} x sin ( x ) - e ^{x} x cos ( x )}{20} + \frac{x ^{2} + x}{4} + \frac{1}{8} + \frac{1}{40} sin (2 x) + \frac{3}{40} cos (2 x)

y = c_{1} e^{- 2 x} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{3 e ^{x} x sin ( x ) - e ^{x} x cos ( x )}{20} + \frac{x ^{2} + x}{4} + \frac{1}{8} + \frac{1}{40} sin (2 x) + \frac{3}{40} cos (2 x)

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