What is the solution for y^{''}-3y^'+2y=13e^{3x} ?

The solution for y^{''}-3y^'+2y=13e^{3x} is y=c_{1}e^{2x}+c_{2}e^x+13/2 e^{3x}

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y^{''}-3y^'+2y=13e^{3x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 13 e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{13}{2} e^{3 x}

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 13 e^{3 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{13}{2} e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{13}{2} e^{3 x}

y^{^{''}} - y = e^{t} sin (t), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 4 e^{x} cos (x)

y^{^{''}} + 4 y = x cos (2 x)

y^{^{''}} + y = sin (x) + x cos (x)

y^{^{''}} + 2 y^{^{'}} - 5 y = - 2

What is the solution for y^{''}-3y^'+2y=13e^{3x} ?
The solution for y^{''}-3y^'+2y=13e^{3x} is y=c_{1}e^{2x}+c_{2}e^x+13/2 e^{3x}