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

The solution for y^{''}-4y^'+4y=e^xsin(x) is y=c_{1}e^{2x}+c_{2}xe^{2x}+1/2 e^xcos(x)

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{2} e^{x} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{2} e^{x} cos (x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{1}{2} e^{x} cos (x)

y^{^{''}} - 4 y^{^{'}} + 5 y = 10 x^{2} + 4 x + 8

y^{^{''}} + 2 y^{^{'}} + y = (\frac{1}{6}) (e^{- x})

y^{^{''}} + 9 y = 7 x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 0.01 y = - 0.01 \cdot 20

y^{^{''}} + \frac{1}{2} y^{^{'}} + y = 4 sin (\frac{x}{2})

What is the solution for y^{''}-4y^'+4y=e^xsin(x) ?
The solution for y^{''}-4y^'+4y=e^xsin(x) is y=c_{1}e^{2x}+c_{2}xe^{2x}+1/2 e^xcos(x)