What is the solution for y^{''}-3y^'+2y=e^{4t} ?

The solution for y^{''}-3y^'+2y=e^{4t} is y=c_{1}e^{2t}+c_{2}e^t+1/6 e^{4t}

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y^{''}-3y^'+2y=e^{4t}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{4 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{6} e^{4 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{6} e^{4 t}

y = c_{1} e^{2 t} + c_{2} e^{t} + \frac{1}{6} e^{4 t}

What is the solution for y^{''}-3y^'+2y=e^{4t} ?
The solution for y^{''}-3y^'+2y=e^{4t} is y=c_{1}e^{2t}+c_{2}e^t+1/6 e^{4t}