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

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

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

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

y = - cos (t) + 2 sin (t)

Solution steps

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

Solve linear ODE:

y = - cos (t) + 2 sin (t)

y = - cos (t) + 2 sin (t)

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