What is the solution for y^{''}+y=0,y(0)=3,y(1)=-3 ?

The solution for y^{''}+y=0,y(0)=3,y(1)=-3 is y=3cos(t)+((-3-3cos(1))sin(t))/(sin(1))

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

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

y = 3 cos (t) + \frac{( - 3 - 3 cos ( 1 ) ) sin ( t )}{sin ( 1 )}

Solution steps

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

Solve linear ODE:

y = 3 cos (t) + \frac{( - 3 - 3 cos ( 1 ) ) sin ( t )}{sin ( 1 )}

y = 3 cos (t) + \frac{( - 3 - 3 cos ( 1 ) ) sin ( t )}{sin ( 1 )}

y^{^{''}} - 12 y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = 16

y^{^{''}} - π^{2} y = 0

y^{^{'''}} - 19 y^{^{''}} + 110 y^{^{'}} - 200 y = 0

4 y^{^{''}} + 16 y = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 4

What is the solution for y^{''}+y=0,y(0)=3,y(1)=-3 ?
The solution for y^{''}+y=0,y(0)=3,y(1)=-3 is y=3cos(t)+((-3-3cos(1))sin(t))/(sin(1))