First order PDE problem $yu_{x} - xu_{y} = x^2$

Question

I am a beginner to PDE, trying to solve $$yu_{x} - xu_{y} = x^2$$ using characteristic line, first I have \begin{align*} \frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2} \end{align*} Then I get the characteristic line is given by $$C = \frac{1}{2}y^2 - \frac{1}{2}x^2$$

Next, I solve the first term and third term, I have \begin{align*} \frac{du}{x^2} &= \frac{dx}{y} \\ du &= \frac{x^2}{y}dx \end{align*} Here is my problem, that is we can not integrate right-hand side without eliminating the variable $y$, but if we try to replace $y$ in terms of $x$ and $C$, the result does not look integrable and really messy.

Can someone help me, please? Thanks!

Answer 1

\begin{align*} \frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2} \end{align*} From the first two ratios, $$\frac{dx}{y} = -\frac{dy}{x} \implies x~dx~+~y~dy~=0 $$ Integrating, $~x^2+y^2=c^2~\tag1$where $~c~$ is integrating constant.

From the last two ratios, $$ -\frac{dy}{x} = \frac{du}{x^2} \implies du=-x~dy\implies du=-\sqrt{c^2-y^2~}~dy \qquad\text{[using equation $(1)$]}$$ Integrating, $$u=-\dfrac 12~y\sqrt{c^2-y^2}~-~\dfrac {c^2}{2} \sin^{-1}\dfrac yc~+~d\qquad\text{(using direct formula) }$$ $$u=-\dfrac 12~xy~-~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)~+~d\qquad\text{[using equation $(1)$]}$$where $d$ is integrating constant.

Hence the general solution is of the form$$u=-\dfrac 12~xy~-~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)~+~\phi\left(\sqrt{x^2+y^2~}\right)$$where $~\phi~$ is arbitrary function of $~x,~y~$.

or, in the form of$$f\left(\sqrt{x^2+y^2}~,~u~+~\dfrac 12~xy~+~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)\right)=0$$where $~f~$ is arbitrary function of $~x,~y~$.

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Integration formula :

$$\int\sqrt{a^2-x^2~}~dx=\dfrac 12~x\sqrt{a^2-x^2~}+\dfrac{a^2}{2}~\sin^{-1}\left(\dfrac{x}{a}\right)+c$$ where $~c~$ is a constant of integration.

Answer 2

$$yu_x-xu_y=x^2$$ Charpit-Lagrange system of characteristic ODEs: $$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{x^2}$$ A first characteristic equation comes from $\frac{dx}{y}=\frac{dy}{-x}$ $$x^2+y^2=c_1$$

A second characteristic equation comes from $\frac{dx}{\sqrt{c_1-x^2}}=\frac{du}{x^2}$

$du=\frac{x^2dx}{\sqrt{c_1-x^2}}\quad\implies\quad u=\int\frac{x^2dx}{\sqrt{c_1-x^2}}=-\frac{xy}{2}+\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right)+c_2$ $$u+\frac{xy}{2}-\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) =c_2$$ The solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is : $$u+\frac{xy}{2}-\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) =F(x^2+y^2)$$ $$\boxed{u(x,y)=-\frac{xy}{2}+\frac{x^2+y^2}{2}\tan^{-1}\left(\frac{y}{x}\right) +F(x^2+y^2)}$$ $F$ is an arbitrary function.

The function $F$ has to be determined according to some boundary condition which is missing in the wording of the question.