Solving PDE $yu_x - xu_y + x^2 - y^2 = 0$ with method of characteristics

I am struggling to understand one 'trick' that is used in the solution of the ODE's from method of characteristics, which currently doesn't make any sense to me:

13. Solve the 1st order PDE $$yu_x - xu_y + x^2 - y^2 = 0$$ Solution: The characteristic equations are $$\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{y^2-x^2}$$ Comparing the 1st and 2nd ratios, we have $$\frac{dx}{y} = \frac{dy}{-x} \quad\implies\quad x^2+y^2 = c_1$$ Next, we multiply the 1st ratio by $$y$$ and the 2nd ratio by $$x$$ and then add them to obtain $$\frac{y dx + x dy}{y^2-x^2} = \frac{du}{y^2-x^2} \quad\implies\quad u-xy = c_2$$

I understand the logic of multiplying 1st and 2nd ratios by $$y$$ and $$x$$ respectively, but how is it possible to add them together in such manner and, even more, equate the result with the third one.

Any help or explanation of at least some part of it would be highly appreciated!

• Thanks for the editing suggestions, I'm new here, so didn't figure out how site works completely, sorry for that – apavl0v May 1 at 16:33
• To understand this tricks, you must have to read some good books like "Elements of Partial Differential Equations" by I. N. Sneddon. – nmasanta May 1 at 16:49

The "trick" is a well-known property in fraction calculus.

If two fractions are equal : $$\frac{A}{B}=\frac{C}{D}\:,$$ they are equal to any fractions on the form : $$\frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}\:,$$ with any coefficients (constants or not) $$c_1$$ and $$c_2$$ not both nul.

In your question : $$A=dx\quad;\quad B=y\quad;\quad C=dy\quad;\quad D=-x\quad;\quad c_1=y\quad;\quad c_2=x.$$

NOTE :

The property is easy to prove.

$$\frac{c_1A+c_2C}{c_1B+c_2D}=\frac{A}{B}\left(\frac{c_1+c_2\frac{C}{A}}{c_1+c_2\frac{D}{B}}\right)=\frac{A}{B}\left(\frac{c_1+c_2\frac{D}{B}}{c_1+c_2\frac{D}{B}}\right)=\frac{A}{B}$$ because $$\frac{C}{A}=\frac{D}{B}$$ .

• Thank you so much! – apavl0v May 1 at 17:27
• Don't mention it. – JJacquelin May 1 at 17:39