What is the solution for y^{''}-4y=8e^{3x} ?

The solution for y^{''}-4y=8e^{3x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+8/5 e^{3x}

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y^{''}-4y=8e^{3x}

y^{^{''}} - 4 y = 8 e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{8}{5} e^{3 x}

Solution steps

y^{''} (x) - 4 y = 8 e^{3 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{8}{5} e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + \frac{8}{5} e^{3 x}

x^{^{''}} + 25 x = 32 cos (3 t)

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

2 y^{^{''}} + y^{^{'}} + 3 y = 3 cos (3 t) + 2 sin (3 t)

y^{^{''}} - y = x^{2}, y (0) = - 3, y^{^{'}} (0) = 5

x^{^{''}} - 6 x^{^{'}} + 13 x = 2197 t^{2}

What is the solution for y^{''}-4y=8e^{3x} ?
The solution for y^{''}-4y=8e^{3x} is y=c_{1}e^{2x}+c_{2}e^{-2x}+8/5 e^{3x}