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

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

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

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

y = - sin (t)

Solution steps

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

Solve linear ODE:

y = - sin (t)

y = - sin (t)

2 x^{^{''}} + 10 x = 0

2 y^{^{'''}} - 6 y^{^{''}} = 0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = - 3, y^{^{'}} (0) = \frac{1}{3}

2 y^{^{''}} + 3 y^{^{'}} + \frac{2}{5} y = 0

3 y^{^{''}} + 48 y = 0, y (0) = 8, y^{^{'}} (0) = - 33

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