What is the solution for y^{''}-2y^'-3y=2e^{5t} ?

The solution for y^{''}-2y^'-3y=2e^{5t} is y=c_{1}e^{3t}+c_{2}e^{-t}+1/6 e^{5t}

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y^{''}-2y^'-3y=2e^{5t}

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{5 t}

y = c_{1} e^{3 t} + c_{2} e^{- t} + \frac{1}{6} e^{5 t}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 3 y = 2 e^{5 t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- t} + \frac{1}{6} e^{5 t}

y = c_{1} e^{3 t} + c_{2} e^{- t} + \frac{1}{6} e^{5 t}

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

(D^{2} - 4 D + 4) y = 8 (x^{2} + e^{2 x} + sin (2 x))

y^{^{''}} + y = cos (x - \frac{π}{2}), y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} + 5 y^{^{'}} + y = 4 t^{2} + 8 t + 8

What is the solution for y^{''}-2y^'-3y=2e^{5t} ?
The solution for y^{''}-2y^'-3y=2e^{5t} is y=c_{1}e^{3t}+c_{2}e^{-t}+1/6 e^{5t}