What is the solution for y^{''}+4y^'+4y=4sin(2t),y(0)=0,y^'(0)=1 ?

The solution for y^{''}+4y^'+4y=4sin(2t),y(0)=0,y^'(0)=1 is y= 1/2 e^{-2t}+2te^{-2t}-1/2 cos(2t)

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y^{''}+4y^'+4y=4sin(2t),y(0)=0,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 sin (2 t), y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{2} e^{- 2 t} + 2 t e^{- 2 t} - \frac{1}{2} cos (2 t)

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 4 sin (2 t)

Solve linear ODE:

y = \frac{1}{2} e^{- 2 t} + 2 t e^{- 2 t} - \frac{1}{2} cos (2 t)

y = \frac{1}{2} e^{- 2 t} + 2 t e^{- 2 t} - \frac{1}{2} cos (2 t)

y^{^{''}} + 4 y^{^{'}} + 8 y = 4 e^{- 2 x}

y^{^{''}} - 4 y = 10 e^{- 2 x}

y^{^{''}} + 2 y^{^{'}} + 10 y = x, y (0) = 0, y^{^{'}} (0) = - 1

y^{^{''}} + 17 y = e^{6 x}

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

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