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

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

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

2 y^{^{''}} - 3 y^{^{'}} - 2 y = 0, y (0) = 2, y^{^{'}} (0) = - \frac{7}{2}

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

Solution steps

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

Solve linear ODE:

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

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

10 x^{^{''}} + 30 x^{^{'}} + 20 x = 0

y^{^{''}} + \frac{5}{2} y^{^{'}} + y = 0

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

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} - 4 y = 0

y^{^{''}} + (e + 1) y^{^{'}} + ey = 0

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