Find a differential equation for a given solution I've been facing a problem since days till now but I could not find out an answer.
I have a Cauchy-problem like this: $x'(t)=sin(x(t))$ with initial condition $x(0)=\frac{\pi}{2}$.
Now I define this solution $y(t)=\cos(t)$. The question is: find $y'(t)$ such that $\cos(t)$ is a solution.
I've tried to resolve the $x'(t)$ with the separation of variables but the solution was too complex to obtain $y'(t)$.
Another possibility I thought was to verify that $y(t)$ was actually a solution of $x'(t)$ but it was not.
Last thing I tried was to derive both left and right side of $y(t)$ to obtain $y'(t)$ on the left side and $-\sin(t)$ on the rightside. Anyway I want to get something like $-\sin(y(t))$ and not only in function of $t$ theoretically.
Can  you help me find out a solution please?
Thanks in advance.
 A: After explanation I understand that the problem is to find a function $f$ such that $y(t)=\cos t$ is a solution of
$$y'(t)=f(y(t)).$$
Would $f(x)=-\sqrt{1-x^2}$ be acceptable? In general, if the function $y$ is given then $f$ can be chosen as the composition of $y'$ with the inverse function of $y.$
A: Have a look at the first Cauchy's problem you solved,
$$x'(t)=sin(x(t)),$$
 initial condition
$$x(o)=\frac{\pi}{2}$$
As you've done, the solution of this differential equation is $$x(t)=-cos(x(t))+C,$$where C is a constant
Then, taking care of the initial condition we easily find $C=\frac{\pi+2}{2}$.
Finally, we can use this Cauchy's problem to solve the new one in the following way,
$$y'(t)=-sin(y(t)),$$
and an initial condition
because then we'll obtain $y(t)=cos(y(t))+C$
The main point here is defining the initial condition, if we establish $y(0)=0$, then we'll get $C=0$ and thus $y(t)=cos(y(t))$, the solution you are looking for. However this initial conditions is not the same as in the first problem, Are u happy with this solution? Do you need the first initial condition?
