What is the solution for y^{''}-16y=48t-24e^{-4t},y(0)=0,y^'(0)=24 ?

The solution for y^{''}-16y=48t-24e^{-4t},y(0)=0,y^'(0)=24 is y=3e^{4t}-3e^{-4t}-3t+3e^{-4t}t

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y^{''}-16y=48t-24e^{-4t},y(0)=0,y^'(0)=24

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 24

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

Solution steps

y^{''} (t) - 16 y = 48 t - 24 e^{- 4 t}

Solve linear ODE:

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

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

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

y^{^{''}} + y = x e^{x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 24 e^{t} + 36 t^{2}

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

y^{^{''}} - 2 y^{^{'}} + 2 = 0

What is the solution for y^{''}-16y=48t-24e^{-4t},y(0)=0,y^'(0)=24 ?
The solution for y^{''}-16y=48t-24e^{-4t},y(0)=0,y^'(0)=24 is y=3e^{4t}-3e^{-4t}-3t+3e^{-4t}t