What is the solution for y^{''}-2y^'=30e^{-3t},y(0)=1,y^'(0)=0 ?

The solution for y^{''}-2y^'=30e^{-3t},y(0)=1,y^'(0)=0 is y=3e^{2t}-4+2e^{-3t}

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y^{''}-2y^'=30e^{-3t},y(0)=1,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} = 30 e^{- 3 t}, y (0) = 1, y^{^{'}} (0) = 0

y = 3 e^{2 t} - 4 + 2 e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = 3 e^{2 t} - 4 + 2 e^{- 3 t}

y = 3 e^{2 t} - 4 + 2 e^{- 3 t}

2 y^{^{''}} + a y^{^{'}} + a y = e^{- 2 x} + x

y^{^{''}} + 2 y^{^{'}} + 2 y = 4 e^{- x} sec^{3} (x)

y^{^{''}} + 36 y = 36 sec^{2} (6 t)

y^{^{''}} - 3 y^{^{'}} = - 4 - 6 x

y^{^{''}} + y^{^{'}} - 2 y = 1 + t, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for y^{''}-2y^'=30e^{-3t},y(0)=1,y^'(0)=0 ?
The solution for y^{''}-2y^'=30e^{-3t},y(0)=1,y^'(0)=0 is y=3e^{2t}-4+2e^{-3t}