What is the solution for y^{''}-2y^'=30e^{-3t} ?

The solution for y^{''}-2y^'=30e^{-3t} is y=c_{1}e^{2t}+c_{2}+2e^{-3t}

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y^{''}-2y^'=30e^{-3t}

y^{^{''}} - 2 y^{^{'}} = 30 e^{- 3 t}

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

Solution steps

y^{''} (t) - 2 y^{'} (t) = 30 e^{- 3 t}

Solve linear ODE:

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

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

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 + e^{2 t}

y^{^{''}} + 12 y^{^{'}} + 37 y = 4 H (t)

y^{^{''}} - y^{^{'}} + 441 y = 21 sin (21 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 4 e^{x} sin (2 x) + 5 x

What is the solution for y^{''}-2y^'=30e^{-3t} ?
The solution for y^{''}-2y^'=30e^{-3t} is y=c_{1}e^{2t}+c_{2}+2e^{-3t}