# Show that $[y+xf(x^2+y^2)]dx+[yf(x^2+y^2)-x]dy=0$ is not an exact differential equation

The criterion for exactness says that, given $$M,N \in C^1(D)$$, a differential equation $$Mdx+Ndy=0$$ is exact if and only if $$M_y=N_x$$.

We know that

$$M_y =1 + x \frac{\partial}{\partial y} f(x^2+y^2) \\ N_x = -1 + y \frac{\partial}{\partial x} f(x^2+y^2)$$

Hence, if $$M_y=N_x$$,

$$2 = y\frac{\partial}{\partial x} f(x^2+y^2) - x \frac{\partial}{\partial y} f(x^2+y^2)$$

But I don't how to proceed (or how to interpret this identity, really). If $$f$$ is a one-variable function, do these partials mean anything?

• Hint: use the chain rule to see that $\frac{\partial}{\partial x}f(x^2 + y^2) = 2x\cdot f'(x^2 + y^2)$, with $2x$ being the partial w.r.t. $x$ of the two-variable function $x^2 + y^2$ inside the one-variable function $f$. – Andrew Oct 12 '19 at 0:58

$$M_y =1 + x \frac{\partial}{\partial y} f(x^2+y^2)$$ $$N_x = -1 + y \frac{\partial}{\partial x} f(x^2+y^2)$$ Let $$X(x,y)=x^2+y^2$$
$$\frac{\partial}{\partial x} f(x^2+y^2)=\frac{\partial}{\partial x} f(X)=f'(X)\frac{\partial X}{\partial x}=2xf'(X)$$
$$\frac{\partial}{\partial y} f(x^2+y^2)=\frac{\partial}{\partial y} f(X)=f'(X)\frac{\partial X}{\partial y}=2yf'(X)$$
$$M_y =1 + x \frac{\partial}{\partial y} f(x^2+y^2)=1+x\big(2yf'(X) \big) = 1+2xyf'(X)$$ $$N_x = -1 + y \frac{\partial}{\partial x} f(x^2+y^2)=-1+y\big(2xf'(X) \big) =-1+2xyf'(X)$$ Since $$1+2xyf'(X)\neq -1+2xyf'(X)\quad\implies\quad M_y\neq N_x$$