What is the solution for y^{''}-2y^'-3y=4e^{5x} ?

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

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y^{''}-2y^'-3y=4e^{5x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{5 x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 4 e^{5 x}

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 1

y^{^{''}} + 3 y^{^{'}} = 2 x - 3

y^{^{''}} + 2 y^{^{'}} + y = 3 x e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 6 = 0

y^{^{'''}} - y = 1

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