What is the solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 ?

The solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 is y=(1+13/5)e^{t/2}-13/5 e^{-2t}

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2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 7

y = (1 + \frac{13}{5}) e^{\frac{t}{2}} - \frac{13}{5} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = (1 + \frac{13}{5}) e^{\frac{t}{2}} - \frac{13}{5} e^{- 2 t}

y = (1 + \frac{13}{5}) e^{\frac{t}{2}} - \frac{13}{5} e^{- 2 t}

What is the solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 ?
The solution for 2y^{''}+3y^'-2y=0,y(0)=1,y^'(0)=7 is y=(1+13/5)e^{t/2}-13/5 e^{-2t}