What is the solution for 2y^{''}+5y^'-18y=0 ?

The solution for 2y^{''}+5y^'-18y=0 is y=c_{1}e^{2t}+c_{2}e^{-(9t)/2}

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2y^{''}+5y^'-18y=0

2 y^{^{''}} + 5 y^{^{'}} - 18 y = 0

y = c_{1} e^{2 t} + c_{2} e^{- \frac{9 t}{2}}

Solution steps

2 y^{''} (t) + 5 y^{'} (t) - 18 y = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- \frac{9 t}{2}}

y = c_{1} e^{2 t} + c_{2} e^{- \frac{9 t}{2}}

What is the solution for 2y^{''}+5y^'-18y=0 ?
The solution for 2y^{''}+5y^'-18y=0 is y=c_{1}e^{2t}+c_{2}e^{-(9t)/2}