What is the solution for y^{''}+4y^'+8y=24,y^'(0)=0,y(0)=0 ?

The solution for y^{''}+4y^'+8y=24,y^'(0)=0,y(0)=0 is y=e^{-2t}(-3cos(2t)-3sin(2t))+3

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y^{''}+4y^'+8y=24,y^'(0)=0,y(0)=0

y^{^{''}} + 4 y^{^{'}} + 8 y = 24, y^{^{'}} (0) = 0, y (0) = 0

y = e^{- 2 t} (- 3 cos (2 t) - 3 sin (2 t)) + 3

Solution steps

y^{''} (t) + 4 y^{'} (t) + 8 y = 24

Solve linear ODE:

y = e^{- 2 t} (- 3 cos (2 t) - 3 sin (2 t)) + 3

y = e^{- 2 t} (- 3 cos (2 t) - 3 sin (2 t)) + 3

y^{^{''}} + 18 y^{^{'}} - y = t^{- 1} e^{t}

D B = 10 lo g_{10} (\frac{0.1}{1 \cdot 1 0 ^{- 12}})

y^{^{''}} - 8 y^{^{'}} + 16 y = 32 t + 64, y (0) = 3, y^{^{'}} (0) = - 3

y^{^{''}} + 4 y^{^{'}} + 40 y = 12.5 sin (6 t)

y^{^{''}} + 3 y^{^{'}} - 28 y = - 96 t + 56 t^{2}

What is the solution for y^{''}+4y^'+8y=24,y^'(0)=0,y(0)=0 ?
The solution for y^{''}+4y^'+8y=24,y^'(0)=0,y(0)=0 is y=e^{-2t}(-3cos(2t)-3sin(2t))+3