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I am trying to solve an exercise with partial derivatives. I have no issue with the derivation part, but once you do that you are supposed to solve a system of equations to find the solution. I already did a few exercises of this kind. However, I found one with a particular system of equations that I can not figure out how to solve. Multiplying one equation by something to delete one variable and solve the other one does not seem to work. I need to know the value of $\frac{\partial x}{\partial u}$ and $\frac{\partial y}{\partial u}$.

Said system of equations is:

$ \left\{ \begin{aligned} \left(2x+y\right)\frac{\partial x}{\partial u}+x\frac{\partial y}{\partial u}=-2 \\ y\frac{\partial x}{\partial u}+\left(x-2y\right)\frac{\partial y}{\partial u}=-1 \\ \end{aligned} \right. $

The answers are supposed to be: $\frac{\partial x}{\partial u}=\frac{4y-x}{2\left(x^2-2xy-y^2\right)}$ and $\frac{\partial y}{\partial u}=\frac{y-2x}{2\left(x^2-2xy-y^2\right)}$

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  • $\begingroup$ think about $v=xy$. $\endgroup$ Feb 21, 2019 at 22:29
  • $\begingroup$ Can you show your work, even if you don't think it worked? This is just solving two equations in two unknowns, $\partial{x}/\partial{u}$ and $\partial{y}/\partial{u}$. $\endgroup$
    – rogerl
    Feb 21, 2019 at 22:30

1 Answer 1

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The system of equations can be written as $$ \begin{pmatrix} 2x+y & x \\ y & x-2y \\ \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial u} \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \end{pmatrix} $$

Using the formula for the inverse of a $2\times 2$ matrix, we get $$ \begin{pmatrix} \frac{\partial x}{\partial u} \\ \frac{\partial y}{\partial u} \end{pmatrix} = \frac{1}{(2x+y)(x-2y)-yx} \begin{pmatrix} x-2y & -x \\ -y & 2x+y \\ \end{pmatrix} \begin{pmatrix} -2 \\ -1 \end{pmatrix} \\ = \frac{1}{2x^2-4xy-2y^2} \begin{pmatrix} (x-2y)(-2) + (-1)(-x) \\ (-y)(-2) + (2x+y)(-1) \end{pmatrix} \\ \frac{1}{2(x^2-2xy-y^2)} \begin{pmatrix} 4y - x \\ y - 2x \end{pmatrix} $$

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  • $\begingroup$ You are brilliant Sir. That's exactly what I was looking for. Thank you very much $\endgroup$
    – John Smith
    Feb 23, 2019 at 17:55

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