What is the solution for 4y^{''}+9y=0,y(0)=2,y^'(0)=3 ?

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

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4y^{''}+9y=0,y(0)=2,y^'(0)=3

4 y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x e^{- a x^{2}})

\frac{d}{d x} (a (2 e^{- 3 x} x - 3 e^{- 3 x} x^{2}))

\frac{\partial}{\partial x} (- sin (4 x))

f (x) = x^{2} + 2 x + 3

\int 4 \frac{x}{( x + 2 ) ^{(\frac{1}{4})}} d x

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