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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

5 y^{^{''}} + y^{^{'}} = - 4 x, y (0) = 0

x^{^{''}} + 2 x = 1

x^{^{''}} + 2 x = 6

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 2 y = 2 x^{2} - 3 x + 6

y^{^{''}} + 10 y^{^{''}} + 33 y^{^{'}} + 36 y = 4 t e^{- 3 t}

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