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

The solution for y^{''}-4y^'+4y=2x-sin(2x) is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x+1)/2-1/8 cos(2x)

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

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

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x + 1}{2} - \frac{1}{8} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x + 1}{2} - \frac{1}{8} cos (2 x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + \frac{x + 1}{2} - \frac{1}{8} cos (2 x)

y^{^{''}} - y = 3 x e^{x} cos (2 x)

y^{^{''}} + 4 y^{^{'}} + 5 y = 5

4 y^{^{''}} + 4 y^{^{'}} + y = \frac{5 e ^{- \frac{t}{2}}}{t ^{2} + 1}

y^{^{''}} - 5 y^{^{'}} + 6 y = - 3 sin (2 x)

y^{^{''}} + 4 y^{^{'}} + 3 y = 18

What is the solution for y^{''}-4y^'+4y=2x-sin(2x) ?
The solution for y^{''}-4y^'+4y=2x-sin(2x) is y=c_{1}e^{2x}+c_{2}xe^{2x}+(x+1)/2-1/8 cos(2x)