What is the solution for y^{''}+36y=0,y(pi)=0,y(0)=0 ?

The solution for y^{''}+36y=0,y(pi)=0,y(0)=0 is y=c_{2}sin(6t)

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y^{''}+36y=0,y(pi)=0,y(0)=0

y^{^{''}} + 36 y = 0, y (π) = 0, y (0) = 0

y = c_{2} sin (6 t)

Solution steps

y^{''} (t) + 36 y = 0

Solve linear ODE:

y = c_{2} sin (6 t)

y = c_{2} sin (6 t)

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

y^{^{'''}} - y^{^{''}} - 56 y^{^{'}} = 0

y^{^{''}} + y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 1

4 y^{^{''}} - 16 y^{^{'}} + 17 y = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y (0) = 3

What is the solution for y^{''}+36y=0,y(pi)=0,y(0)=0 ?
The solution for y^{''}+36y=0,y(pi)=0,y(0)=0 is y=c_{2}sin(6t)