What is the solution for 4y^{''}+9y=0 ?

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

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4y^{''}+9y=0

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

y = c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})

Solution steps

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

Solve linear ODE:

y = c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})

y = c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2})

\int \frac{4 x ^{3}}{x ^{4} + 7} d x

f (x) = 2 x^{2} + 5 x

\int 2 t d t

- x^{2} + 4 x + 1, x + 1

\int \frac{( y - 3 )}{y} d y

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