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

The solution for y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=1 is y=e^{-4t}+5e^{-4t}t

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} - 6 y = 0

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

y^{^{''}} + y = 0, y (\frac{π}{3}) = 2, y^{^{'}} (\frac{π}{3}) = - 4

y^{^{''}} + 4 y^{^{'}} + 16 y = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 32 y = 0

What is the solution for y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+8y^'+16y=0,y(0)=1,y^'(0)=1 is y=e^{-4t}+5e^{-4t}t