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

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

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

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

y = cos (t)

Solution steps

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

Solve linear ODE:

y = cos (t)

y = cos (t)

\frac{d ^{3} y}{d x ^{3}} + \frac{d ^{2} y}{d x ^{2}} = 0

64 y^{^{''}} + 144 y^{^{'}} + 81 y = 0

y^{^{'''}} + 3 y^{^{''}} - 22 y^{^{'}} - 24 y = 0

2 y^{^{''''}} + 3 y^{^{'''}} + 2 y^{^{''}} + 6 y^{^{'}} - 4 y = 0

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 3

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