What is the solution for y^{''}+4y^'+68y=0,y(0)=7 ?

The solution for y^{''}+4y^'+68y=0,y(0)=7 is y=e^{-2t}(7cos(8t)+c_{2}sin(8t))

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y^{''}+4y^'+68y=0,y(0)=7

y^{^{''}} + 4 y^{^{'}} + 68 y = 0, y (0) = 7

y = e^{- 2 t} (7 cos (8 t) + c_{2} sin (8 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 68 y = 0

Solve linear ODE:

y = e^{- 2 t} (7 cos (8 t) + c_{2} sin (8 t))

y = e^{- 2 t} (7 cos (8 t) + c_{2} sin (8 t))

2 x^{^{''}} + 32 x = 0, x (0) = 0, x^{^{'}} (0) = 0

y^{^{''}} + y = 0, y (0) = 11, y (0) = 33

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

y^{^{'''}} + 2 y^{^{''}} - 11 y^{^{'}} - 52 y = 0

\frac{d ^{2} f}{d x ^{2}} + 64 f (x) = 0

What is the solution for y^{''}+4y^'+68y=0,y(0)=7 ?
The solution for y^{''}+4y^'+68y=0,y(0)=7 is y=e^{-2t}(7cos(8t)+c_{2}sin(8t))