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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (0.75 - 2 x sin (0.1 x))

y^{^{''}} + 25 y = 10 sin (5 x)

x \to 0 lim (ln (3 - x))

\frac{s}{s ^{2} + 8}

e^{x}, 5

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