What is the solution for 2y^{''}+3y^'+9y=0 ?

The solution for 2y^{''}+3y^'+9y=0 is y=e^{-(3t)/4}(c_{1}cos((3sqrt(7)t)/4)+c_{2}sin((3sqrt(7)t)/4))

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2y^{''}+3y^'+9y=0

2 y^{^{''}} + 3 y^{^{'}} + 9 y = 0

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{3 7 t}{4}) + c_{2} sin (\frac{3 7 t}{4}))

Solution steps

2 y^{''} (t) + 3 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{3 7 t}{4}) + c_{2} sin (\frac{3 7 t}{4}))

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{3 7 t}{4}) + c_{2} sin (\frac{3 7 t}{4}))

\frac{( x ^{4} - 5 x ^{3} + x )}{x ^{2}}

f (x) = (3 x^{2} - 2 x - 1)^{\frac{3}{2}}

\int \frac{38}{38 + e ^{x}} d x

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What is the solution for 2y^{''}+3y^'+9y=0 ?
The solution for 2y^{''}+3y^'+9y=0 is y=e^{-(3t)/4}(c_{1}cos((3sqrt(7)t)/4)+c_{2}sin((3sqrt(7)t)/4))