What is the solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=1 is y=2e^{-2t}+5te^{-2t}

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y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = 2 e^{- 2 t} + 5 t e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = 2 e^{- 2 t} + 5 t e^{- 2 t}

y = 2 e^{- 2 t} + 5 t e^{- 2 t}

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 3

k x^{^{''}} + a x = 0

\frac{d ^{2} x}{d t ^{2}} + a^{2} x = 0

y^{^{''}} - 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{'''}} + 5 y^{^{''}} + y^{^{'}} - 75 y = 0, y (0) = e^{3 x}

What is the solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+4y^'+4y=0,y(0)=2,y^'(0)=1 is y=2e^{-2t}+5te^{-2t}