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

The solution for y^{''}+3y^'+2y=0,y(0)=1,y^'(0)=3 is y=5e^{-t}-4e^{-2t}

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

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

y = 5 e^{- t} - 4 e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = 5 e^{- t} - 4 e^{- 2 t}

y = 5 e^{- t} - 4 e^{- 2 t}

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

9 y^{^{''}} - 4 y = 0, y^{1} (0) = e \frac{2 x}{3}

3 y^{^{''}} = 4 y^{^{'}}

y^{^{''}} + y^{^{'}} = - cy

324 y^{^{''}} + 36 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = 7

What is the solution for y^{''}+3y^'+2y=0,y(0)=1,y^'(0)=3 ?
The solution for y^{''}+3y^'+2y=0,y(0)=1,y^'(0)=3 is y=5e^{-t}-4e^{-2t}