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

The solution for y^{''}-4y=xe^x+cos(2x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^xx)/3-2/9 e^x-1/8 cos(2x)

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

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

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{2}{9} e^{x} - \frac{1}{8} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{2}{9} e^{x} - \frac{1}{8} cos (2 x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{3} - \frac{2}{9} e^{x} - \frac{1}{8} cos (2 x)

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (5 t)

y^{^{''}} + y^{^{'}} - 2 y = 2 x, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 8 y^{^{'}} + 16 y = 1 - x + e^{4 x}

y^{^{''}} - y^{^{'}} - 6 y = csc (x)

y^{^{'''}} - 2 y^{^{''}} - 5 y^{^{'}} + 6 y = e^{- 2 x}

What is the solution for y^{''}-4y=xe^x+cos(2x) ?
The solution for y^{''}-4y=xe^x+cos(2x) is y=c_{1}e^{2x}+c_{2}e^{-2x}-(e^xx)/3-2/9 e^x-1/8 cos(2x)