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

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

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

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

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

Solution steps

y^{''} (t) + y - 1 = 0

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

y^{^{''}} - y^{^{'}} = 4 x e^{x}

d, d (q) = \frac{300}{0.4 q + 8}

y^{^{''}} + 4 y^{^{'}} + 3 y = 1 - H (t - 2) + H (t - 6)

2 y^{^{''}} + y^{^{'}} - 3 y = e^{x} - x

y^{^{''}} + 4 y = \frac{1}{sin ( 2 t )}

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