On pose $r=\sqrt{x^2+y^2}$ et $\theta=\Arctan\left(\frac{y}{x}\right)$ de sorte que $x=r\cos\theta$ et $y=r\sin\theta$ . On pose $f(x,y)=f(r\cos\theta,r\sin\theta)=g(r,\theta)$ . On sait que $\frac{\partial r}{\partial x}=\cos\theta$ , $\frac{\partial r}{\partial y}=\sin\theta$ , $\frac{\partial \theta}{\partial x}=-\frac{\sin\theta}{r}$ , $\frac{\partial \theta}{\partial y}=\frac{\cos\theta}{r}$ $\begin{align*}\ensuremath x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}=r\cos\theta\left(\cos\theta\frac{\partial g}{\partial r}-\frac{\sin\theta}{r}\frac{\partial g}{\partial \theta}\right)+r\sin\theta\left(\sin\theta\frac{\partial g}{\partial r}+\frac{\cos\theta}{r}\frac{\partial g}{\partial \theta}\right)=r\frac{\partial g}{\partial r}, \end{align*}$ puis

$\begin{align*}\ensuremath \forall(x,y)\in D,\;x\frac{\partial f}{\partial x}(x,y)&+ y\frac{\partial f}{\partial y}(x,y)=\sqrt{x^2+y^2}\Leftrightarrow\forall r>0,\;r\frac{\partial g}{\partial r}(r,\theta)=r\Leftrightarrow\forall r>0,\;\frac{\partial g}{\partial r}(r,\theta)=1\\ &\Leftrightarrow\exists \varphi\;\text{de classe}\;C^1\;\text{sur}\;\left]-\frac{\pi}{2},\frac{\pi}{2}\right[/\;\forall(r,\theta)\in]0,+\infty[\times\left]-\frac{\pi}{2},\frac{\pi}{2}\right[,\;g(r,\theta)=r+\varphi(\theta)\\ &\Leftrightarrow\exists \varphi\;\text{de classe}\;C^1\;\text{sur}\;\left]-\frac{\pi}{2},\frac{\pi}{2}\right[/\;\forall(x,y)\in D,\;f(x,y)=\sqrt{x^2+y^2}+\varphi\left(\Arctan\frac{y}{x}\right)\\ &\Leftrightarrow\exists \psi\;\text{de classe}\;C^1\;\text{sur}\;\Rr/\;\forall(x,y)\in D,\;f(x,y)=\sqrt{x^2+y^2}+\psi\left(\frac{y}{x}\right). \end{align*}$

exo7 5562