# Thread: How to solve the equation???

1. ## How to solve the equation???

Dear All,

Need your helps desperately! I have no idea how to solve this equation, or even don't know if the solution exists at all. Can you help me? Or relate me to books or topics for this type of equations?

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Let f(x,y) be known, find function v(x)=? such that:

v[f(x,y)]=v(x)*(df/dx)+w(y)*(df/dy)

Where w(y) can be selected arbitrarily so that v(x) does exist.

Note: v(x) is function of x only, w(y) is function of y only and can be choosed by usere to ensure existence of v(x). df/dx and df/dy are partial derivatives of x and y respectively.
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Thanks very much for your help indeed!!!

2. Originally Posted by WangTaoSG
Dear All,

Need your helps desperately! I have no idea how to solve this equation, or even don't know if the solution exists at all. Can you help me? Or relate me to books or topics for this type of equations?

************************************************** *****
Let f(x,y) be known, find function v(x)=? such that:

v[f(x,y)]=v(x)*(df/dx)+w(y)*(df/dy)

Where w(y) can be selected arbitrarily so that v(x) does exist.

Note: v(x) is function of x only, w(y) is function of y only and can be choosed by usere to ensure existence of v(x). df/dx and df/dy are partial derivatives of x and y respectively.
************************************************** *****

Thanks very much for your help indeed!!!
If this has a soultion in general we would have for a known function $g(x)$ that there would be a $v(x)$ such that:

$v(g(x))=v(x) \frac{d}{dx}g(x)$

but let $g(x)=x^2$, then:

$v(x^2)=v(x)(2x)$

Then:

$v(1)=2v(1).$

(or suppose $v(x)$ has a power series repersentation valid in some interval, then look at the consequence of $v(x^2)=2xv(x)$)

CB

3. ## Which means solution needs not exist always then...

the question becomes:
(1) under what condition such solution does exist and
(2) how to solve it

Really appreciate you example here, but wonder if there are books or topics that teach about these types of equations?

Thanks very much!!