Prove that differential equation $y' = 1 + \sqrt[5]{y-x}$ has exactly one solution Given differential equation $y' = 1 + \sqrt[5]{y-x}$
Prove that through every point of region $(x-2)^2 + (y-4)^2=1$ passes exactly one solution of given equation  
In my opinion, we have to use Cauchy theorem on several regions, union of which gives initial region, but i'm stuck
 A: Let us consider the function $f$ of the variable $(x,y)$, defined on $\Bbb R^2$ by
$$
f(x,y)= 1+(y-x)^{1/5}\ .
$$
Fix $\delta >0$.
We would like to apply Picard-Lindelöf for $f$ and any point $(x_0,y_0)$ in the open region
$$
D_\delta:=\{\ (x,y)\in\Bbb R^2\ :\ \delta < y-x\ \}\ ,
$$
which is the half-plane containing $(2,4)$ with boundary parallel and
at distance  $\delta$ from the line $y=x$. The given function is continuous in $x$. It remains to insure $f$ is uniformly Lipschitz continuous in the variable $y$. For this, we build the differential $f'_y$ of $f$ w.r.t. the variable $y$, it is
$$
f'_y(x,y)
=
\frac 15 (y-x)^{-4/5}
=
\frac 1{5(y-x)^{4/5}}
\ .
$$
Observe that this expression is bounded, since the distance from the closure of $D_\delta$ to the line $x=y$ is $\delta>0$. The reason to restrict to $_\delta$ was to avoid $(3,3)$:
$B_\delta$ to the line $x=y$, dan_fulea" />
We have $y-x\ge \delta$, so
$$
\|f'_y\|\le \frac 1{5\delta^{4/5}}\ .
$$
The conditions for Picard-Lindelöf for $f$ on $D_\delta$ are thus satisfied, we have a unique solution locally "through" each point $(x_0,y_0)$ in $D_\delta$. And the regions $D_\delta$ are covering the given ball around $(2,4)$ with radius one, except for the point $(3,3)$. So we need to consider this point separately. The equation to be solved is
$$
(y-x)'=(y-x)^{1/5}\ ,\ y(3)=3\ .
$$
One solution is $y=x$. We have to check if  there is no other.
The problem begins here, we find all solutions below.

Consider  an other solution $y:(3-a,3+a)\to \Bbb R$. (We want to get the solution or a contradiction soon.)
Let $b\ge 0$ be maximal so that $y=x$ on $[3-b,3+b]$. The case $b>0$ is not possible, since then $y(3\pm b)=3\pm b$, $y=x$ is still a local solution, and with a similar argument as above (possibly for a half-plane on the other side) we obtain a contradiction to the uniqueness.
So $b=0$.
And we have $y\ne x$ on some strictly monotone sequence inside $(3-a,3+a)$ converging to $3$. Then locally around a point $x_0$ from this sequence we obtain successively:
$$
\begin{aligned}
(y-x)'&=(y-x)^{1/5}\ ,\\
(y-x)'(y-x)^{-1/5}&=1\ ,\\
\Big(\ (y-x)^{4/5}\ \Big)'&=\frac 45\ ,\\
(y-x)^{4/5}&=\frac 45x+C(x_0)\ .
\end{aligned}
$$
Here $C(x_0)$ is a constant function living on the maximal interval inside $\{x\ :\ y(x)\ne x\}$ that contains $x_0$.
If $x_0<3$, then the L.H.S. is $>0$, so the R.H.S. is $>0$ and it remains so locally to the right of $x_0$ on the interval $(x_0,3)$. So also in the point $3$. But there the L.H.S. is $(y(3)-3)^{4/5}=0$. A contradiction. This implies that we have $y(x)=x$ for $x\le 3$.
Let us consider now the case $x_0>3$. Then again, on the interval $(x_0,3+a)$ the R.H.S. remains $>0$, so the formula remains true as the unique possibility to extend the solution. This implies
$$
(y-x)^{4/5}=\frac 45 x+C\ ,\ x\in[3,3+a)\ ,
$$
for a suitable constant $C$. (We can define $C$ as the limit of $C(x_0)$ for $x_0\searrow 3$.) This and $y(3)=3$ determines $C$, so that we have
$$
(y-x)^{4/5}=\frac 45(x-3)  ,\ x\in[3,3+a)\ ,
$$
We extract the formula for $y$ on $[3,3+a)$ and
consider the only possibility
$$
y(x)=
\begin{cases}
x &\text{ for }x\le 3\ ,\\
x+\left(\frac 45\right)^{5/4}(x-3)^{5/4} &\text{ for }x\ge 3\ .
\end{cases}
$$
We observe that the two branches do fit together into a $\mathcal C^1$ function.
So the claimed property fails for the point $(3,3)$.
