Two functions $u = u(x,y)$ and $v = v(x,y)$ are defined by the system of equations: $$u-v=x-y$$ $$yu-xv=1$$

The problem asks for the partial derivatives $\frac {\partial u} {\partial x},~ \frac {\partial u} {\partial y},~ \frac {\partial v} {\partial x},~\frac {\partial v} {\partial y}$ and the added picture below is showing my solution using a Jacobian matrix. We have received the our professor's solution and he solved the system manually without the Jacobian by solving the system simultaneously. It seems I have got the correct answer for both partials with respect to x. But the partials with respect to y are seemingly not.

In his solution $\frac {\partial u}{\partial y}=\frac {u-x}{x+y}$ and $\frac {\partial v}{\partial y}=\frac{u+y}{x+y}$

I don't understand why my method worked for the partials with respect to x, but not with respect to y. How can that be? If both are wrong, then fine, there must be some mistake, but one is correct one is not with the same method? What am I missing?

Thanks in advance for any help on this.

$$\\$$ enter image description here


1 Answer 1


I think your solution is correct.

Solving the system 'manually' (as you said the professor did?), let's take the partial derivatives with respect to $y$ and then solve the system for $u_y$ and $v_y$ (using subscripts for partial derivatives), we get: $$\left\{ \begin{array}{rcl} u-v&=&x-y \\ yu-xv&=&1 \end{array} \right. \implies \left\{ \begin{array}{rcl} u_y-v_y&=&-1 \\ u+yu_y-xv_y&=&0 \end{array} \right. \iff \ldots \iff \left\{ \begin{array}{rcl} u_y&=&\tfrac{u-x}{x-y} \\ v_y&=&\tfrac{u-y}{x-y} \end{array} \right.$$ This matches your calculations. You also expect the denominator to contain $x-y$ rather than $x+y$ since, as you found, the Jacobian is non-zero for $x \ne y$.


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