What is the solution for y^{'''}-2y^{''}+2y^'=1+xe^xe ?

The solution for y^{'''}-2y^{''}+2y^'=1+xe^xe is

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y^{'''}-2y^{''}+2y^'=1+xe^xe

y^{^{'''}} - 2 y^{^{''}} + 2 y^{^{'}} = 1 + x e^{x} e

y = c_{1} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{1}{2} (x + e^{1 + x} x - e^{1 + x}) + e^{x} cos (x) \cdot \int - \frac{( sin ( x ) + cos ( x ) ) ( 1 + e ^{x + 1} x )}{2 e ^{x}} d x + e^{x} sin (x) \cdot \int \frac{( cos ( x ) - sin ( x ) ) ( 1 + e ^{x + 1} x )}{2 e ^{x}} d x

Solution steps

y^{'''} (x) - 2 y^{''} (x) + 2 y^{'} (x) = 1 + x e^{x} e

Solve linear ODE:

y = c_{1} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{1}{2} (x + e^{1 + x} x - e^{1 + x}) + e^{x} cos (x) \cdot \int - \frac{( sin ( x ) + cos ( x ) ) ( 1 + e ^{x + 1} x )}{2 e ^{x}} d x + e^{x} sin (x) \cdot \int \frac{( cos ( x ) - sin ( x ) ) ( 1 + e ^{x + 1} x )}{2 e ^{x}} d x

y = c_{1} + e^{x} (c_{2} cos (x) + c_{3} sin (x)) + \frac{1}{2} (x + e^{1 + x} x - e^{1 + x}) + e^{x} cos (x) \cdot \int - \frac{( sin ( x ) + cos ( x ) ) ( 1 + e ^{x + 1} x )}{2 e ^{x}} d x + e^{x} sin (x) \cdot \int \frac{( cos ( x ) - sin ( x ) ) ( 1 + e ^{x + 1} x )}{2 e ^{x}} d x

y^{^{''}} + 6 y^{^{'}} + 34 y = 68, y^{^{'}} (0) = 7, y (0) = 5

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

(d^{4} + 5 d^{2} + 4) y = 2 cos (x)

2 y^{^{''}} - y^{^{'}} - y = 5 x e^{2 x}

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

What is the solution for y^{'''}-2y^{''}+2y^'=1+xe^xe ?
The solution for y^{'''}-2y^{''}+2y^'=1+xe^xe is