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I have a question: $$p^{2}x+q^{2}y = z.$$

I formed the Charpit auxiliary equation as follows

$$ \frac{\mathrm{d}x}{2px} = \frac{\mathrm{d}y}{2py} = \frac{\mathrm{d}z}{2(p^2x + q^2y)} = \frac{\mathrm{d}p}{p-p^2} = \frac{\mathrm{d}q}{q-q^2}.$$

After forming the equation I was unable to solve further (I applied everything I was taught). So I did some research (checked several books, and on the website as well) and found the next step to be

$$\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}. \tag{1}$$

After which I was able to solve. But I can not understand how they derived the relation (1). Can someone explain what method they applied here?

Disclaimer: The course we are being taught is engineering mathematics.

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Charpit's equation: $$\boxed{\frac{dx}{2px} = \frac{dy}{\color{red}{2qy}} = \frac{dz}{2(p^2x + q^2y)} = \frac{dp}{p-p^2} = \frac{dq}{q-q^2}}$$ We have from Charpit's equation $$\frac{dy}{2qy} =\frac{dq}{q-q^2}$$ Rearranging terms $$(q-q^2){dy} =2qy{dq}$$ $$q{dy} =2qy{dq}+q^2dy$$ $$\text { (1) }\frac{dy}{qy} =\frac {2qy{dq}+q^2dy}{q^2y}$$ We have also from Charpit's equation that : $$\frac{dx}{2px} = \frac{dp}{p-p^2}$$ $${dx}{(p-p^2)} = 2px{dp}$$ $${pdx} = 2px{dp}+p^2dx$$ $$\text { (2) }\frac {dx}{px} = \frac {2px{dp}+p^2dx} {p^2x}$$ We use a third equality from Charpit's equation $$\frac {dx}{2px}=\frac{dy}{2qy} \implies \frac {dx}{px}=\frac{dy}{qy} $$ Therefore $$\boxed{\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}}$$ That you can easily integrate...

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Note that the middle term is equal to $\frac{dz}{2z}$ and use it as the reference value. Then $$ \frac{p^2dx}{p^2x}=\frac{pdz}{z},\quad \frac{2px\,dp}{p^2x}=\frac{2dp}{p}=\frac{(1-p)\,dz}{z} \implies \frac{d(p^2x)}{p^2x}=\frac{dz}{z} $$ and similarly for $q^2y$.


Some other obvious simple steps are to recognize the decoupled nature of $$ \frac{dz}{2z}=-\frac{d(1/p)}{1/p-1}=-\frac{d(1/q)}{1/q-1} $$ which integrates to $$ \frac12\ln|z|=c_1-\ln|1/p-1|=c_2-\ln|1/q-1| \\\iff\\ \sqrt{|z|}=C_1\frac{p}{1-p}=C_2\frac{q}{1-q} $$ Then insert $$ p=\frac{\sqrt{|z|}}{C_1+\sqrt{|z|}},\qquad q=\frac{\sqrt{|z|}}{C_2+\sqrt{|z|}} $$ into the first three relations $$ \frac{dx}{2xp}=\frac{dy}{2yq}=\frac{dz}{2z} $$ which should then be easy to integrate.


Or alternatively, combine the $x$ and $p$ fractions to $$ \frac{dx}{x}=\frac{2dp}{1-p}\implies x(1-p)^2=a, $$ and similarly $$ \frac{dy}{y}=\frac{2dq}{1-q}\implies y(1-q)^2=b. $$

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  • $\begingroup$ Can you explain as to how we got the relation (1) mentioned in the question. Also what do you mean by decoupled nature? Please explain, knowing that I have a undergraduate understanding of PDEs. $\endgroup$ – Mohammed Arshaan Mar 3 '18 at 12:08
  • $\begingroup$ Also what method is being applied here to transform the auxiliary equations? As far as I can see they are not simple addition, subtraction, multiplication or division operations... $\endgroup$ – Mohammed Arshaan Mar 3 '18 at 12:09
  • $\begingroup$ The first transformations are exactly that, simple arithmetic operations. Note that $p-p^2=p(1-p)$. As both expressions in (1) are equal to $\frac{dz}{z}$, they are also equal to each other. $\endgroup$ – LutzL Mar 3 '18 at 12:16
  • $\begingroup$ Or did you question $d(p^2x)=p^2\,dx+2px\,dp$? $\endgroup$ – LutzL Mar 3 '18 at 12:16
  • $\begingroup$ No. What I mean is that in relation (1) (marked in my question). How did we get the following relation: $$\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}. \tag{1}$$ $\endgroup$ – Mohammed Arshaan Mar 3 '18 at 12:18

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