What is the solution for 9y^{''}+24y^'-33y=0 ?

The solution for 9y^{''}+24y^'-33y=0 is y=c_{1}e^t+c_{2}e^{-(11t)/3}

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9y^{''}+24y^'-33y=0

9 y^{^{''}} + 24 y^{^{'}} - 33 y = 0

y = c_{1} e^{t} + c_{2} e^{- \frac{11 t}{3}}

Solution steps

9 y^{''} (t) + 24 y^{'} (t) - 33 y = 0

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- \frac{11 t}{3}}

y = c_{1} e^{t} + c_{2} e^{- \frac{11 t}{3}}

\frac{d ^{2} y}{d x ^{2}} = (4 y)

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = 1

y^{^{'''}} + 3 y^{^{''}} + 4 y^{^{'}} + 12 y = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y^{^{''}} + 4 y = 0, y (\frac{π}{4}) = - 2

What is the solution for 9y^{''}+24y^'-33y=0 ?
The solution for 9y^{''}+24y^'-33y=0 is y=c_{1}e^t+c_{2}e^{-(11t)/3}