Find $b$ and $c$ such that for all $x$ and $y$, $z = x^2 + bxy + cy^2$ and $\partial z/\partial x = \partial z/\partial y$

Question

Find $b$ and $c$ such that for all $x$ and $y$,

$$z = x^2 + bxy + cy^2\quad \text{and}\quad\frac{\partial z}{\partial x} =\frac{\partial z}{\partial y}$$

@DonAntonio Sir after you get (2−b)x=(2c−b)y, how can you get the b and c? — iak, Dec 14 '13 at 16:58

score 2 · Answer 1 · answered Dec 14 '13 at 16:41

2

Hints:

$$\frac{\partial z}{\partial x}=2x+by\\ \frac{\partial z}{\partial y}=2cy+bx$$

So you have to solve for all $\;x,y\;$ :

$$2x+ by=bx+2cy\iff(2-b)x=(2c-b)y\;\;\ldots\ldots$$

answered Dec 14 '13 at 16:41

DonAntonio

Sir I'm still confused how to get b and c – iak Dec 14 '13 at 16:46
Sir after you get (2−b)x=(2c−b)y, how can you get the b and c? – iak Dec 14 '13 at 16:58
The secrete is in "for all $;x,y\in\Bbb R;$ " , so for example choose $;y=0;$ and check what happens... – DonAntonio Dec 14 '13 at 16:59
Sir I got zero but i'm still confused – iak Dec 14 '13 at 17:02
@iak, what happens in the last equality if $;y=0;,;;x\neq 0;$ ?? Common, this is differential equations, you must know this! – DonAntonio Dec 14 '13 at 17:02

1 Answers1