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

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

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

linear

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

y = - e^{- 3 t} - 1

Solution steps

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

Solve linear ODE:

y = - e^{- 3 t} - 1

y = - e^{- 3 t} - 1

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 6, y^{^{'}} (0) = 2

y^{^{''}} + 121 y = 0, y (0) = sin (k t)

2 y^{^{''}} + 11 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 3, y^{^{'}} (0) = 16

\frac{3}{4} x^{^{''}} + 192 x = 0, x (0) = - \frac{1}{4}, x^{^{'}} (0) = 0

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