What is the solution for y^{''}-8y^'-4y-2=2xe^xsin(x) ?

The solution for y^{''}-8y^'-4y-2=2xe^xsin(x) is

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y^{''}-8y^'-4y-2=2xe^xsin(x)

y^{^{''}} - 8 y^{^{'}} - 4 y - 2 = 2 x e^{x} sin (x)

y = c_{1} e^{2 (2 + 5) x} + c_{2} e^{2 (2 - 5) x} + e^{4 x + 25 x} \cdot \int \frac{5 e ^{- 4 x - 25 x} ( e ^{x} x sin ( x ) + 1 )}{10} d x + e^{4 x - 25 x} \cdot \int - \frac{5 e ^{25 x - 4 x} ( e ^{x} x sin ( x ) + 1 )}{10} d x

Solution steps

y^{''} (x) - 8 y^{'} (x) - 4 y - 2 = 2 x e^{x} sin (x)

Solve linear ODE:

y = c_{1} e^{2 (2 + 5) x} + c_{2} e^{2 (2 - 5) x} + e^{4 x + 25 x} \cdot \int \frac{5 e ^{- 4 x - 25 x} ( e ^{x} x sin ( x ) + 1 )}{10} d x + e^{4 x - 25 x} \cdot \int - \frac{5 e ^{25 x - 4 x} ( e ^{x} x sin ( x ) + 1 )}{10} d x

y = c_{1} e^{2 (2 + 5) x} + c_{2} e^{2 (2 - 5) x} + e^{4 x + 25 x} \cdot \int \frac{5 e ^{- 4 x - 25 x} ( e ^{x} x sin ( x ) + 1 )}{10} d x + e^{4 x - 25 x} \cdot \int - \frac{5 e ^{25 x - 4 x} ( e ^{x} x sin ( x ) + 1 )}{10} d x

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 e^{2 x}, y (0) = 1, y^{^{'}} (0) = 6

y^{^{''}} - 6 y^{^{'}} + 9 y = 5 t^{2}, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 7

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 6

y^{^{''}} - 8 y^{^{'}} + 16 y = 16 x + 5

What is the solution for y^{''}-8y^'-4y-2=2xe^xsin(x) ?
The solution for y^{''}-8y^'-4y-2=2xe^xsin(x) is