What is the solution for y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=-2 ?

The solution for y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=-2 is y=e^{-4t}+2e^{-4t}t

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y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=-2

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y = e^{- 4 t} + 2 e^{- 4 t} t

Solution steps

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

Solve linear ODE:

y = e^{- 4 t} + 2 e^{- 4 t} t

y = e^{- 4 t} + 2 e^{- 4 t} t

y^{^{''}} + 16 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - 12

y^{^{''}} + 3 y^{^{'}} + 7 y = 0

y^{^{''}} + y^{^{'}} = y

y^{^{'''}} + 2 y^{^{''}} - 19 y^{^{'}} - 20 y = 0

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 14 y = 0

What is the solution for y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=-2 ?
The solution for y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=-2 is y=e^{-4t}+2e^{-4t}t