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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 16 y = e^{4 x}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{8 t}

\frac{d ^{2} y}{d x ^{2}} - k \frac{d y}{d x} = x e^{x}

- 2 y^{^{''}} + y = e^{- 0.2 x}

2 y^{^{''}} - 6 y^{^{'}} + 3 y = 4 sin (x) + 6 x

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