- #1

- 287

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xU

_{x}+yU

_{y}=nu

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

^{n}f(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 x

^{n}come into play

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- Thread starter chaotixmonjuish
- Start date

- #1

- 287

- 0

xU

and its solution (according to the back) is: u(x,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 x

- #2

- 287

- 0

- #3

MathematicalPhysicist

Gold Member

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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?

- #4

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

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The form of the question suggests that you should use the method of characteristics (see http://www.stanford.edu/class/math220a/handouts/firstorder.pdf" [Broken] 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).

xU_{x}+yU_{y}=nu

and its solution (according to the back) is: u(x,y)=x^{n}f(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 x^{n}come into play

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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