What is the solution for 4y^{''}+9y=0,y(0)=0,y^'(0)=1 ?

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

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4y^{''}+9y=0,y(0)=0,y^'(0)=1

4 y^{^{''}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

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