Partial Differential Equations Lagrange method

Question

In the following question: $\frac{dx}{x^2}=\frac{dx}{y^2}$=$\frac{dz}{z(x+y)}$

I have to approach the solution in two different methods however both my answer conflicts.

Method 1: Subtracting 1 and 2 and equating with 3rd

$x^{-2}dx$=$y^{-2}dy$

which gives on integrating:

$\frac{-1}{x}+\frac{1}{y}$=$constant$

Method 2:

$\frac{zdx/x}{x}$=$\frac{zdy/y}{y}$=$\frac{dz}{x+y}$

now adding one and two subtracting from third

$\frac{dx}{x}$+$\frac{dy}{y}$=$\frac{dz}{z}$

which gives on integrating,

$xy/z=(constant)$

how ever both are different,in my book they subtract one and two then use the property $x^2-y^2=(x+y)(x-y)$ to cancel it with x+y which gives another constant however I think there are supposed to be only two solutions is there something wrong Iam doing please guide me through

Answer 1

Usig $x^2-y^2=(x+y)(x-y)$ we get the integral $\dfrac{x-y}{z}=c_3$. It says that the expression $\dfrac{x-y}{z}$ is constant along the characteristics.

You got two other expressions, $\dfrac{-1}{x}+\dfrac{1}{y}=c_1$ and $\dfrac{xy}{z}=c_2$, two constants along the characteristics. The point is that we can set $c_3$ from $c_1$ and $c_2$ alone:

$\dfrac{x-y}{xy}=c_1$ and $xy=zc_2$, then

$\dfrac{x-y}{zc_2}=c_1$ or $\dfrac{x-y}{z}=c_1c_2$

Meaning that $c_3=c_1c_2$. So, $c_3$ is not an independent constant. There are only two independent solutions.

Answer 2

Probably there is a typo in your system of ODEs ($dx$ instead of $dy$) in the second fraction.

Sorry I cannot understand what you are doing because they are not enough steps edited when going from one equation to the next. Also you write "Subtracting 1 and 2 and equating with 3rd" without making clear what are 1, 2 and 3. If 1 is $\frac{dx}{x^2}$ and if 2 is $\frac{dy}{y^2}$ substracting gives $0$ since both are equal.

Moreover I don't understand why you are surprized because "Method 1" and "Method 2" give different results. Fortunately they give different results ! Each one gives a characteristic equation wich are different one from the other.

A first characteristic equation is $$\frac{1}{y}-\frac{1}{x}=c_1$$ A second characteristic equation is $$\frac{z}{xy}=c_2$$ I suppose that the PDE is $$x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=(x+y)z$$ which leads to the Charpit-Lagrange system of ODEs : $$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$$ Thanks to the two above characteristic equations one can get to the general solution of the PDE on implicit form $c_2=F(c_1)$ and on explicit form : $$z(x,y)=xy\,F\left(\frac{1}{y}-\frac{1}{x}\right)$$ where $F$ is an arbitrary function (to be determined according to some conditions not given in the wording of the question).

VERIFICATION :

Let $\chi=\frac{1}{y}-\frac{1}{x}\quad;\quad z(x,y)=xyF(\chi)\quad$ and with symbol $\quad F'=\frac{dF}{d\chi}$. $$\frac{\partial z}{\partial x}= yF+xy\frac{\partial \chi}{\partial x}=yF+ \frac{y}{x}F'$$ $$\frac{\partial z}{\partial y}= xF+xy\frac{\partial \chi}{\partial y}=xF- \frac{x}{y}F'$$ $$x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=x^2\left(yF+\frac{y}{x}F'\right)+y^2\left(xF-\frac{x}{y}F'\right)$$ $$x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=x^2yF+y^2xF=(x+y)xyF$$ $$x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=(x+y)z$$ This is exactly the original PDE. Thus $z=xy\,F\left(\frac{1}{y}-\frac{1}{x}\right)$ is solution of the PDE.