What is the solution for y^{''}-4y=2x-3cos(x)+e^{-x} ?

The solution for y^{''}-4y=2x-3cos(x)+e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-2x}-x/2+3/5 cos(x)-1/3 e^{-x}

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y^{''}-4y=2x-3cos(x)+e^{-x}

y^{^{''}} - 4 y = 2 x - 3 cos (x) + e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{x}{2} + \frac{3}{5} cos (x) - \frac{1}{3} e^{- x}

Solution steps

y^{''} (x) - 4 y = 2 x - 3 cos (x) + e^{- x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{x}{2} + \frac{3}{5} cos (x) - \frac{1}{3} e^{- x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} - \frac{x}{2} + \frac{3}{5} cos (x) - \frac{1}{3} e^{- x}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{1 + e ^{t}}

y^{^{''}} - y^{^{'}} + \frac{1}{4} y = 8 + e^{\frac{x}{2}}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{2 x} + 4 x - 12

y^{^{''}} + y - 5 = e^{2 x} - 3 + cos (3 x)

y^{^{''}} - 4 y = 2 e^{3 x}

What is the solution for y^{''}-4y=2x-3cos(x)+e^{-x} ?
The solution for y^{''}-4y=2x-3cos(x)+e^{-x} is y=c_{1}e^{2x}+c_{2}e^{-2x}-x/2+3/5 cos(x)-1/3 e^{-x}