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

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

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

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

y = - cos (t) + c_{2} sin (t)

Solution steps

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

Solve linear ODE:

y = - cos (t) + c_{2} sin (t)

y = - cos (t) + c_{2} sin (t)

2 y^{^{''}} - 6 y^{^{'}} + 8 y = 0

y^{^{''''}} - 4 y^{^{''}} = 0

a \cdot y^{^{''}} + b \cdot y = 0

y^{^{''}} + 4 y^{^{'}} - 60 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = 4

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