What is the solution for y^{''}+y=0,y(pi/2)=0,y^'(pi/2)=1 ?

The solution for y^{''}+y=0,y(pi/2)=0,y^'(pi/2)=1 is y=-cos(t)

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y^{''}+y=0,y(pi/2)=0,y^'(pi/2)=1

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

y = - cos (t)

Solution steps

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

Solve linear ODE:

y = - cos (t)

y = - cos (t)

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

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

y^{^{''}} + (a - 1) y^{^{'}} + b y = 0

y^{^{'''}} + 3 y^{^{''}} = 0

8 y^{^{''}} - 9 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = \frac{1}{8}

What is the solution for y^{''}+y=0,y(pi/2)=0,y^'(pi/2)=1 ?
The solution for y^{''}+y=0,y(pi/2)=0,y^'(pi/2)=1 is y=-cos(t)