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?

  • 1
    $\begingroup$ 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 $f$. $\endgroup$ – 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 $1+2xyf'(X)\neq -1+2xyf'(X)\quad\implies\quad M_y\neq N_x$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.