What is the solution for 3y^{''}+2y^'-8y=0,y(0)=-6,y^'(0)=-18 ?

The solution for 3y^{''}+2y^'-8y=0,y(0)=-6,y^'(0)=-18 is y=-9e^{(4t)/3}+3e^{-2t}

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3y^{''}+2y^'-8y=0,y(0)=-6,y^'(0)=-18

3 y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 6, y^{^{'}} (0) = - 18

y = - 9 e^{\frac{4 t}{3}} + 3 e^{- 2 t}

Solution steps

3 y^{''} (t) + 2 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = - 9 e^{\frac{4 t}{3}} + 3 e^{- 2 t}

y = - 9 e^{\frac{4 t}{3}} + 3 e^{- 2 t}

y^{^{''}} + 42 y^{^{'}} + 8 y = 0

\frac{d ^{3} y}{d x ^{3}} - 21 \frac{d y}{d x} - 20 y = 0

10 y^{^{''}} - 3 y^{^{'}} = 0

y^{^{''}} + 3 y^{^{'}} + 8 y = 0

\frac{d ^{2} x}{d t ^{2}} = x

What is the solution for 3y^{''}+2y^'-8y=0,y(0)=-6,y^'(0)=-18 ?
The solution for 3y^{''}+2y^'-8y=0,y(0)=-6,y^'(0)=-18 is y=-9e^{(4t)/3}+3e^{-2t}