Why is the method of characteristics used to solve this PDE?

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SUMMARY

The method of characteristics is essential for solving the partial differential equation (PDE) given by xUx + yUy = nu. The solution is u(x,y) = xnf(y/x), where f(y/x) is derived from the characteristic equation c = y/x. The discussion highlights the importance of verifying solutions by substituting them back into the original equation. It also emphasizes that the method of characteristics is a systematic approach rather than a trial-and-error process.

PREREQUISITES
  • Understanding of partial differential equations (PDEs)
  • Familiarity with the method of characteristics
  • Knowledge of logarithmic functions and their properties
  • Ability to verify solutions by substitution
NEXT STEPS
  • Study the method of characteristics in detail using resources like the Stanford handout on first-order PDEs
  • Explore the derivation of solutions for different types of PDEs
  • Practice solving PDEs using the method of characteristics with various initial conditions
  • Investigate the implications of the solution u(x,y) = xnf(y/x) in physical contexts
USEFUL FOR

Mathematicians, engineering students, and anyone studying or working with partial differential equations, particularly those interested in the method of characteristics and its applications.

chaotixmonjuish
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I borrowed this book form the library to get a heads up on what i would be doing in PDEs and it had this equation

xUx+yUy=nu

and its solution (according to the back) is: u(x,y)=xnf(y/x)

I did figure out the f(y/x) part:

dx/x=dy/y
ln(x)+c=ln(y)
cx+y
c=y/x


how did the xn come into play
 
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My attempt at a solution yielded a yn. I don't believe that is wrong, but I am wondering why they are multiplied together instead of added.
 
You ofcourse can check your or the book's answer by plugging it back to the equation.
I think that there is no easy way here, the mathematician who find these equations knew already what solution he was seeking and just played with its derivatives to get to this equation.
I don't have time to check your work, but have you plugged your answers back to the eqaution to check that it's valid?
 
chaotixmonjuish said:
I borrowed this book form the library to get a heads up on what i would be doing in PDEs and it had this equation

xUx+yUy=nu

and its solution (according to the back) is: u(x,y)=xnf(y/x)

I did figure out the f(y/x) part:

dx/x=dy/y
ln(x)+c=ln(y)
cx+y
c=y/x


how did the xn come into play
The form of the question suggests that you should use the method of characteristics (see http://www.stanford.edu/class/math220a/handouts/firstorder.pdf" for more information). Note that the method you have used above to find the characteristics is not generally correct and it is just by 'chance' that you have arrived at the correct characteristic equation (c=y/x).

The link I provided above should give you everything you need, if not, I suggest reading the preceeding chapter of your book.
 
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