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

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

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y^{''}+9y=0,y^'(pi/2)=0,y(0)=0

y^{^{''}} + 9 y = 0, y^{^{'}} (\frac{π}{2}) = 0, y (0) = 0

y = c_{2} sin (3 t)

Solution steps

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

Solve linear ODE:

y = c_{2} sin (3 t)

y = c_{2} sin (3 t)

y^{^{''}} + y^{^{'}} - 6 y = 0, y (1) = 2 e^{2} - e^{- 3}

(D^{3} - 3 D^{2} + 3 D - 1) y = 0

8 y^{^{''}} - 9 y^{^{'}} + y = 0

y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} - 26 y = 0

y^{^{''}} - 2 y^{^{'}} + 5 y = 0

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