What is the solution for 2y^{''}+5y^'-3y=0,y(0)=22 ?

The solution for 2y^{''}+5y^'-3y=0,y(0)=22 is y=(22-c_{2})e^{t/2}+c_{2}e^{-3t}

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2y^{''}+5y^'-3y=0,y(0)=22

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 22

y = (22 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- 3 t}

Solution steps

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

Solve linear ODE:

y = (22 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- 3 t}

y = (22 - c_{2}) e^{\frac{t}{2}} + c_{2} e^{- 3 t}

What is the solution for 2y^{''}+5y^'-3y=0,y(0)=22 ?
The solution for 2y^{''}+5y^'-3y=0,y(0)=22 is y=(22-c_{2})e^{t/2}+c_{2}e^{-3t}