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

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

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

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

y = e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = e^{- 2 t}

y = e^{- 2 t}

10 y^{^{''''}} + 13 y^{^{''}} + 4 y = 0

y^{^{''}} - 4 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 4, y^{^{'}} (- 1) = 6

\frac{d ^{2} y}{d x ^{2}} = a y

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y^{1} (0) = e^{3} x

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