What is the solution for y^{''}+8y^'+16y=0,y(0)=-4,y^'(0)=21 ?

The solution for y^{''}+8y^'+16y=0,y(0)=-4,y^'(0)=21 is y=-4e^{-4t}+5e^{-4t}t

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y^{''}+8y^'+16y=0,y(0)=-4,y^'(0)=21

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = - 4, y^{^{'}} (0) = 21

y = - 4 e^{- 4 t} + 5 e^{- 4 t} t

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = - 4 e^{- 4 t} + 5 e^{- 4 t} t

y = - 4 e^{- 4 t} + 5 e^{- 4 t} t

\frac{d ^{2}}{d x ^{2}} (x (3 x + 2)^{5})

y = \frac{x}{x ^{2} + 5}

\frac{\partial}{\partial x} (1 - y)

\frac{\partial}{\partial x} (\frac{a}{ln ( x ) - c})

y^{^{'}} + \frac{4 y}{x} = - \frac{4}{x}

What is the solution for y^{''}+8y^'+16y=0,y(0)=-4,y^'(0)=21 ?
The solution for y^{''}+8y^'+16y=0,y(0)=-4,y^'(0)=21 is y=-4e^{-4t}+5e^{-4t}t