Multiple Solutions to an ODE

Question

I want to find the solution to the IVP: $$y' = 2\cos(x)\sqrt{y-1},\;\; y\geq1 $$ with initial condition $y(0)=2$.

I used separation of variables to get the general solution $$y =\left(\sin(x)+C\right)^2 +1.$$

When I use the initial condition to try and find a specific solution, I get that $C = \pm 1$.

Both of these values of $C$ work when subbed back into the original differential equation.

But since the ODE is continuous for all $y\geq 1$ and the $y$ partial derivative is continuous for all $y>1$, Picard's theorem says that there should be a unique solution for the initial condition $y(0)=2$.

Is there any way that I can further test the two solutions to see which one is invalid?

Answer 1

The two solutions \begin{align}y_1(x)&=(\sin x-1)^2+1\\y_2(x)&=(\sin x+1)^2+1\end{align} represent the same function but shifted. That is, $\sin(x+\pi)-1=-(\sin x+1)$ so $y_1$ is $y_2$ shifted by $\pi$.

Answer 2

You need also to ensure that, as per the given equation, $y'(0)=2\cos(0)\sqrt{2-1}=2$, that is, the slope has to be positive. But $$ y_1'(x)=2\cos x(\sin x-1)\implies y_1'(0)=-2. $$

Answer 3

The solution $y=(\sin x+C)^2+1$ is only a solution to your differential equation for $C> 0$. In the process of solving the equation, you end up with something like: $$\sqrt{y-1}=\sin x+C$$ Since this is the positive square root we need $\sin x+C\ge 0$ at all points where $y$ is defined; if we want $y$ to be defined on a neighborhood around $0$ then we need $C>0$. (For $C\ge 1$ the solution will be valid for all $x$.)

Note that the function $y=(\sin x+C)^2+1$ is always a solution to the differential equation $$(y')^2=4\cos^2x\,(y-1)$$ for all $C$, both positive and negative. You've shown that this latter DE with initial condition $y(0)=2$ indeed doesn't have a unique solution.