What is the solution for y^{''}+25y=0,y^'(pi)=0,y^'(0)=0 ?

The solution for y^{''}+25y=0,y^'(pi)=0,y^'(0)=0 is y=c_{1}cos(5t)

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y^{''}+25y=0,y^'(pi)=0,y^'(0)=0

y^{^{''}} + 25 y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 0

y = c_{1} cos (5 t)

Solution steps

y^{''} (t) + 25 y = 0

Solve linear ODE:

y = c_{1} cos (5 t)

y = c_{1} cos (5 t)

10 y^{^{''}} + 6 y^{^{'}} + 10.9 y = 0

y^{^{''}} (s) - 3 y^{^{'}} (s) + y (s) 3 = 0 es

y^{^{''''}} - k^{2} y^{^{''}} = 0

y^{^{''}} + \frac{2}{5} y^{^{'}} + \frac{1}{25} y = 0, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} + \frac{2}{5} y^{^{'}} + \frac{1}{25} y = 0, y (0) = 0, y^{^{'}} (0) = 4

What is the solution for y^{''}+25y=0,y^'(pi)=0,y^'(0)=0 ?
The solution for y^{''}+25y=0,y^'(pi)=0,y^'(0)=0 is y=c_{1}cos(5t)