How to solve the following pde?

Question

How to solve the following PDE?

For an arbitrary continuously differentiable function $f$ , which of the following is a general solution of $\;$ $z(px-qy)=y^2-x^2$?

1)$\;$ $x^2+y^2+z^2=f(xy)$

2)$\;$ $(x+y)^2+z^2=f(xy)$

3)$\;$ $x^2+y^2+z^2=f(y-x)$

4)$\;$ $x^2+y^2+z^2=f((x+y)^2+z^2)$

Here options $1)$$\;$ $2)$ and $ 4)$ are correct but I am getting only first option as answer.

Is there any general method to solve such pdes? Please help because I don't have any teacher who can help me and I am learing pde without any teacher.

Thank you very much for giving me your precious time.

Where is this question from? What are $p$ and $q$? It doesn't appear to be a PDE and I don't see how $f$ has anything to do with the equation you're trying to solve? — charlestoncrabb, Dec 01 '15 at 18:52
@charlestoncrabb You can find this question here question number 98.http://www.csirhrdg.res.in/mathCEN_June2015.pdf — zafran, Dec 01 '15 at 18:54
I know of lagrange method of multiplier, jacobi method so far for this kind of pde. I guess, not sure. Or one where you had to take characteristic curves or something but it's all blurry or something for me. — Someone, Dec 01 '15 at 19:03

doraemonpaul · Accepted Answer · 2015-12-05T15:42:57.887

2

$z(xz_x-yz_y)=y^2-x^2$

$xz_x-yz_y=\dfrac{y^2-x^2}{z}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$

$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$

$\dfrac{dz}{dt}=\dfrac{y^2-x^2}{z}=\dfrac{y_0^2e^{-2t}-e^{2t}}{z}$ , we have $z^2=f(y_0)-y_0^2e^{-2t}-e^{2t}=f(xy)-y^2-x^2$ , i.e. $x^2+y^2+z^2=f(xy)$

$\therefore$ 1) is obviously correct.

But 2) is in fact also correct since $(x+y)^2=x^2+2xy+y^2$ .

From 2), $g((x+y)^2+z^2)=xy$

$\therefore x^2+y^2+z^2=f(g((x+y)^2+z^2))$ , equivalent to $x^2+y^2+z^2=f((x+y)^2+z^2)$

Hence 4) is in fact also correct.

See rule 6 for further details.

edited Dec 05 '15 at 15:42

answered Dec 05 '15 at 15:17

doraemonpaul

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I didn't get how 2) is correct from $(x+y)^2=x^2+2xy+y^2$? Could you please explain? Which rule 6? – zafran Dec 05 '15 at 15:22
1

@zafran $f(xy)-2xy$ is equivalent to $f(xy)$ . – doraemonpaul Dec 05 '15 at 15:29
By the way which rule 6 are you talking about? – zafran Dec 05 '15 at 15:41
1

@zafran the instructions of csirhrdg.res.in/mathCEN_June2015.pdf which you gave in your comment. – doraemonpaul Dec 05 '15 at 15:47

score 0 · Answer 2 · answered May 07 '16 at 10:52

Adding another way of looking at it: Taking Lagrange Multiplier,

$${dx\over zx}={dy\over -zq}={dz\over y^2-x^2}$$

First two gives $xy=c$ and $x\times$ first $+y\times$second=$-z\times$third gives $${(x+y)d(x+y)\over zx^2-zy^2}={-zdz\over zx^2-zy^2}$$ and gives $(x+y)^2+z^2=c$.

That adds (2) and (4)

How to solve the following pde?

2 Answers2