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

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

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

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

y = c_{1} e^{3 t} + c_{2} e^{t} - e^{2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{t} - e^{2 t}

y = c_{1} e^{3 t} + c_{2} e^{t} - e^{2 t}

y^{^{''}} + y^{^{'}} = - 16 cos (3 x)

1 y^{^{''}} + 5 y^{^{'}} + \frac{1}{\frac{1}{6}} y = 2 e^{2 t}

y^{^{''}} + 2 y^{^{'}} - 2 y = 5 sin (x) - 10 x e^{x}

y^{^{''}} - 8 y^{^{'}} + 17 y = - 102, y (0) = - 3, y^{^{'}} (0) = 8

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3} x ln (x)

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