What is the solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=1 is y=e^{-t}-e^{-2t}

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y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=1

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

(4 d^{4} - 8 d^{3} - 4 d^{2} + 8 d) y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 20 y = 0, y (0) = 4, y^{^{'}} (0) = - 15

4 y^{^{''}} + π^{2} y = 0

3 y^{^{'''}} + 2 y^{^{''}} = 0

What is the solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+3y^'+2y=0,y(0)=0,y^'(0)=1 is y=e^{-t}-e^{-2t}