What is the solution for y^{''}+8y^'+32y=0,y^0=1,y^'0=-4 ?

The solution for y^{''}+8y^'+32y=0,y^0=1,y^'0=-4 is y=e^{-4t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}+8y^'+32y=0,y^0=1,y^'0=-4

y^{^{''}} + 8 y^{^{'}} + 32 y = 0, y^{0} (0) = 1, y^{^{'}} 0 = - 4

y = e^{- 4 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 32 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 4 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y^{^{''}} + 7 y^{^{'}} - 12 y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 8 y = 0

y^{^{''}} + y = 0, y (0) = 3, y (0) = 30

y^{^{''}} + 4^{2} y = 0

3 \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = 0

What is the solution for y^{''}+8y^'+32y=0,y^0=1,y^'0=-4 ?
The solution for y^{''}+8y^'+32y=0,y^0=1,y^'0=-4 is y=e^{-4t}(c_{1}cos(4t)+c_{2}sin(4t))