What is the solution for 2y^{''}+3y^'-2y=0,y(0)=-1,y^'(0)=-4 ?

The solution for 2y^{''}+3y^'-2y=0,y(0)=-1,y^'(0)=-4 is y=(-1-7/5)e^{t/2}+7/5 e^{-2t}

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2y^{''}+3y^'-2y=0,y(0)=-1,y^'(0)=-4

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 0, y (0) = - 1, y^{^{'}} (0) = - 4

y = (- 1 - \frac{7}{5}) e^{\frac{t}{2}} + \frac{7}{5} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = (- 1 - \frac{7}{5}) e^{\frac{t}{2}} + \frac{7}{5} e^{- 2 t}

y = (- 1 - \frac{7}{5}) e^{\frac{t}{2}} + \frac{7}{5} e^{- 2 t}

y^{^{'''}} + 3 y^{^{''}} - 4 y^{^{'}} - 12 y = 0

(D^{5} + 8 D^{3} + 16 D) y = 0

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

y^{^{''}} + 2 y^{^{'}} - 2 y = 0

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

What is the solution for 2y^{''}+3y^'-2y=0,y(0)=-1,y^'(0)=-4 ?
The solution for 2y^{''}+3y^'-2y=0,y(0)=-1,y^'(0)=-4 is y=(-1-7/5)e^{t/2}+7/5 e^{-2t}