What is the solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=2 ?

The solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=2 is y=2e^{-t}-2e^{-2t}

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

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

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

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 16 \frac{d y}{d x} + 64 y = 0

y^{^{''}} + 14 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 37 y = 0

2 x^{^{''}} + 3 x^{^{'}} + 4 x = 0

(D^{3} - 3 D^{2} - D + 3) y = 0

What is the solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=2 ?
The solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=2 is y=2e^{-t}-2e^{-2t}