What is the solution for y^{''}+16y^'+64y=0,y(0)=-3,y^'(0)=28 ?

The solution for y^{''}+16y^'+64y=0,y(0)=-3,y^'(0)=28 is y=-3e^{-8t}+4e^{-8t}t

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y^{''}+16y^'+64y=0,y(0)=-3,y^'(0)=28

y^{^{''}} + 16 y^{^{'}} + 64 y = 0, y (0) = - 3, y^{^{'}} (0) = 28

y = - 3 e^{- 8 t} + 4 e^{- 8 t} t

Solution steps

y^{''} (t) + 16 y^{'} (t) + 64 y = 0

Solve linear ODE:

y = - 3 e^{- 8 t} + 4 e^{- 8 t} t

y = - 3 e^{- 8 t} + 4 e^{- 8 t} t

\int_{0}^{8} 1 + \frac{81}{4} x d x

4 x^{2} - 6 x + 2

\frac{d}{d x} (5 sin (x) (\frac{4}{x ^{3}} + 3 x))

arccos (x + 1)

\int_{0}^{1} 1 + t d t

What is the solution for y^{''}+16y^'+64y=0,y(0)=-3,y^'(0)=28 ?
The solution for y^{''}+16y^'+64y=0,y(0)=-3,y^'(0)=28 is y=-3e^{-8t}+4e^{-8t}t