What is the solution for 2y^{''}+5y^'+3y=0,y(0)=3,y^'(0)=-4 ?

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

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2y^{''}+5y^'+3y=0,y(0)=3,y^'(0)=-4

2 y^{^{''}} + 5 y^{^{'}} + 3 y = 0, y (0) = 3, y^{^{'}} (0) = - 4

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

Solution steps

2 y^{''} (t) + 5 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

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

y^{^{''''}} + 6 y^{^{'''}} + 9 y^{^{''}} = 0

y^{^{'''}} + 2 y^{^{''}} - 7 y^{^{'}} + 4 y = 0

y^{^{''}} - 8 y^{^{'}} - 12 y = 0

y^{^{''''}} + 2 y^{^{'''}} + 10 y^{^{''}} + 18 y^{^{'}} + 9 y = 0

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