Prove that there is at most one function sx satisfying s0 0

Prove that there is at most one function s(x) satisfying s(0) = 0, s\'(0) = 1, s\"(x) = -s(x) for all x belongsto R. (Use Taylor\'s Theorem).

Solution:

Recall the Existence and uniqueness theorem of second order ODE y^\'\'(x)+ p(x)y^\'+q(x)y=r(x)

y\'\'(x)+ p(x)y\'+q(x)y=r(x), y(x₀)=y_0,y\'(x₀)=y₀

_{has unique solution on [a,b].}

_{Now, S\"(x)+1.S(x)+0=0 where p(x)=1 and q(x)=0 are continuous on [0,1].}

_{By above threorem there is unique S.}

To find the solution:

Te auxiliary equation is m²=-1 Or m= +i,-i

Therefore the general solution is: S(x)= Acos(x)+Bsin(x), where A,B are arbitrary constants.

Apply initial conditions:

S(0)=0 gives, Acos(0)+Bsin(0)=0. implies,A=0.

So S(x)=Bsin(x). implies, S^\'(x)=Bcos(x).

S^\'(0)= 1 gives, Bcos(0)=1.implies B=1.

Therefore, the unique solution is S(x)=sin(x).

$Prove that there is at most one function s(x) satisfying s(0) = 0, s\'(0) = 1, s\$