What is the solution for 2y^{''}-y^'-3y=0,y(0)=0,y^'(0)=5 ?

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

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2y^{''}-y^'-3y=0,y(0)=0,y^'(0)=5

2 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 5

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + \frac{220}{55} y = 0

y^{^{''}} + 16 y = 0, y (0) = 2, y^{^{''}} (0) = - 2

y^{^{'''}} + y^{^{''}} - 16 y^{^{'}} + 20 y = 0

3 y^{^{''}} + 5 y^{^{'}} + y = 0

y^{^{''}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 1

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