What is the solution for y^{''}-2y^'-3y=8e^{5x} ?

The solution for y^{''}-2y^'-3y=8e^{5x} is y=c_{1}e^{3x}+c_{2}e^{-x}+2/3 e^{5x}

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y^{''}-2y^'-3y=8e^{5x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e^{5 x}

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

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 8 e^{5 x}

Solve linear ODE:

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

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

y^{^{''}} + 8 y = 5 x + 2 e^{- x}

y^{^{''}} + a \frac{d y}{d t} + b y = c sin (g t)

x^{^{''}} + 4 x^{^{'}} + 13 x = 5 cos (2 t)

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

y^{^{''}} - 16 y = 3

What is the solution for y^{''}-2y^'-3y=8e^{5x} ?
The solution for y^{''}-2y^'-3y=8e^{5x} is y=c_{1}e^{3x}+c_{2}e^{-x}+2/3 e^{5x}