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

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

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

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

y = - cos (t) + \frac{( cos ( 1 ) - 1 ) sin ( t )}{sin ( 1 )} + 1

Solution steps

y^{''} (t) = 1 - y

Solve linear ODE:

y = - cos (t) + \frac{( cos ( 1 ) - 1 ) sin ( t )}{sin ( 1 )} + 1

y = - cos (t) + \frac{( cos ( 1 ) - 1 ) sin ( t )}{sin ( 1 )} + 1

(4 d^{2} + 12 d + 9) y = 144 e^{- 3} x

y^{^{''}} + y^{^{'}} = x^{2} + 2 x + 4

y^{^{''}} + 2 y = x + 1

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 = 0

y^{^{'''}} - 6 y^{^{''}} = 4 - sin (x)

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