The differential equation

$xu_x + yu_y =2u$

satisfying the initial condition $y = xg(x)$ , $u = f(x)$ with

  1. $f(x) = 2x $ , $g(x) = 1$ has no solution.

  2. $f(x) = 2x^2 $ , $g(x) = 1$ has infinite number of solutions.

  3. $f(x) = x^3 $ , $g(x) = x$ has a unique solution.

  4. $f(x) = x^4 $ , $g(x) = x$ has a unique solution.

which one(s) of them will be correct?

My attempt :

${dx \over x}$ = ${dy \over y}$ = ${du \over 2u}$

having solved them and putting initial conditions I get $f(x) = x^2 \phi(g(x))$..then what to do ? can anyone please help me out?

I can not understand what has been said here.

Can you anyone please explain to me in simple language?


From the link you gave, we can get the general solution. It is $u=y^2h(y/x)$ with $h$ any single variable smooth function. (You can check it)

1.- if along $y=xg(x)=1$ $u=f(x)=2x$ then $x^2h(1)=2x$ or $h(1)=2/x$ Meaningless, no solutions.

2.- if along $y=xg(x)=x$ $u=f(x)=2x^2$ then $x^2h(1)=2x^2$ or $h(1)=2$ an infinity of solutions as there is an infinity of functions fulfilling the restriction on only one point.

3.- if along $y=xg(x)=x^2$ $u=f(x)=x^3$ then $x^2h(x^2/x)=x^3$ or $h(x)=x$, so,

$h(y/x)=(y/x)$ and $u=y^2(y/x)$. One solution: $u(x,y)=y^3/x$

4.- if along $y=xg(x)=x^2$ $u=f(x)=x^4$ then $x^2h(x^2/x)=x^4$ or $h(x)=x^2$, so,

$h(y/x)=(y/x)^2$ and $u=y^2(y/x)^2$. One solution: $u(x,y)=y^4/x^2$

  • $\begingroup$ Can you please check this question once more and edit your answer accordingly.? Actually I have made some changes....@Rafa Budria $\endgroup$ – INDIAN May 10 '18 at 4:14
  • $\begingroup$ @INDIAN checked. Now the proposed solutions match. $\endgroup$ – Rafa Budría May 10 '18 at 11:15
  • $\begingroup$ so you are saying 3 and 4 are correct?@Rafa Budria $\endgroup$ – INDIAN May 10 '18 at 11:17
  • $\begingroup$ With some qualification, yes: the solution is unique for $\mathbb R^2-\{(0,0)\}$, the function is not defined at $(0,0)$ $\endgroup$ – Rafa Budría May 10 '18 at 11:23
  • $\begingroup$ means???????????? are 3 and 4 correct?@Rafa Budria $\endgroup$ – INDIAN May 10 '18 at 11:25

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