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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

2 y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

x^{^{''}} + x^{^{'}} + 2 x = 0

y^{^{''}} + y^{^{'}} - 20 y = 0, y (0) = 1, y (0) = 22

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