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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?

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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: $u(x,y)=y^4/x^2$

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  • $\begingroup$ 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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