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=-cos(t)

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

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

y = - cos (t)

Solution steps

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

Solve linear ODE:

y = - cos (t)

y = - cos (t)

x^{^{''}} + 2 x^{^{'}} + 10 x = 0, x^{^{'}} (0) = 1, x (0) = 0

x^{^{''}} + 6 x^{^{'}} + 10 x = 0

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 3, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 16 y = 0, y (0) = 5

y^{^{''''}} + 4 y^{^{'''}} + 13 y^{^{''}} + 36 y^{^{'}} + 36 y = 0

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=-cos(t)