What is the solution for 3y^{''}+2y^'=0,y(0)=-6,y^'(0)=-18 ?

The solution for 3y^{''}+2y^'=0,y(0)=-6,y^'(0)=-18 is y=27e^{-(2t)/3}-33

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3y^{''}+2y^'=0,y(0)=-6,y^'(0)=-18

3 y^{^{''}} + 2 y^{^{'}} = 0, y (0) = - 6, y^{^{'}} (0) = - 18

y = 27 e^{- \frac{2 t}{3}} - 33

Solution steps

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

Solve linear ODE:

y = 27 e^{- \frac{2 t}{3}} - 33

y = 27 e^{- \frac{2 t}{3}} - 33

\frac{3}{16} y^{^{''}} + 12 y = 0, y (0) = \frac{2}{3}, y^{^{'}} (0) = - 3

y^{^{''}} + 14 y^{^{'}} + 49 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + (a^{2} - 2) y = 0

y^{^{''''}} - y = 0

25 x^{^{''}} + 5 x^{^{'}} + 204.167 x = 0, x (0) = 28.577

What is the solution for 3y^{''}+2y^'=0,y(0)=-6,y^'(0)=-18 ?
The solution for 3y^{''}+2y^'=0,y(0)=-6,y^'(0)=-18 is y=27e^{-(2t)/3}-33