What is the solution for y^{''}-2y-3y=4e^x-9 ?

The solution for y^{''}-2y-3y=4e^x-9 is y=c_{1}e^{sqrt(5)x}+c_{2}e^{-sqrt(5)x}-e^x+9/5

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y^{''}-2y-3y=4e^x-9

y^{^{''}} - 2 y - 3 y = 4 e^{x} - 9

y = c_{1} e^{5 x} + c_{2} e^{- 5 x} - e^{x} + \frac{9}{5}

Solution steps

y^{''} (x) - 2 y - 3 y = 4 e^{x} - 9

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- 5 x} - e^{x} + \frac{9}{5}

y = c_{1} e^{5 x} + c_{2} e^{- 5 x} - e^{x} + \frac{9}{5}

y^{^{''}} + 25 y = cos (4 x)

y^{^{''}} - y^{^{'}} - 6 y = 69 cos (3 x)

\frac{d ^{3} y}{d x ^{3}} + \frac{d y}{d x} = e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 x + 1

4 x^{^{''}} + x = t^{2} + 2 cos (t)

What is the solution for y^{''}-2y-3y=4e^x-9 ?
The solution for y^{''}-2y-3y=4e^x-9 is y=c_{1}e^{sqrt(5)x}+c_{2}e^{-sqrt(5)x}-e^x+9/5