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

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

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

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

y = 3 e^{- t} - 2 e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = 3 e^{- t} - 2 e^{- 2 t}

y = 3 e^{- t} - 2 e^{- 2 t}

3 y^{^{''}} + 24 y^{^{'}} + 48 y = 0, y (0) = 8, y^{^{'}} (0) = - 33

y^{^{''}} - 4 y^{^{'}} + 29 y = 0, y (0) = 6, y^{^{'}} (0) = - 3

6 y^{^{''}} - 13 y^{^{'}} + 6 y = 0

y^{^{''}} + 5 y^{^{'}} + 8 y = 0

y^{^{'''}} - 19 y^{^{'}} + 30 y = 0

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