The differential equation $xu_x + yu_y =2u$ satisfying the initial condition $y = xg(x)$ , $u = f(x)$ with

Question

The differential equation

$xu_x + yu_y =2u$

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

$f(x) = 2x $ , $g(x) = 1$ has no solution.
$f(x) = 2x^2 $ , $g(x) = 1$ has infinite number of solutions.
$f(x) = x^3 $ , $g(x) = x$ has a unique solution.
$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?

Rafa Budría · Answer 1 · 2018-05-10T11:13:19.287

0

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

edited May 10 '18 at 11:13

answered May 09 '18 at 16:37

Rafa Budría

7,364

Can you please check this question once more and edit your answer accordingly.? Actually I have made some changes....@Rafa Budria – INDIAN May 10 '18 at 04:14
@INDIAN checked. Now the proposed solutions match. – Rafa Budría May 10 '18 at 11:15
so you are saying 3 and 4 are correct?@Rafa Budria – INDIAN May 10 '18 at 11:17
With some qualification, yes: the solution is unique for $\mathbb R^2-{(0,0)}$, the function is not defined at $(0,0)$ – Rafa Budría May 10 '18 at 11:23
means???????????? are 3 and 4 correct?@Rafa Budria – INDIAN May 10 '18 at 11:25
Yes! No more solutions than one! – Rafa Budría May 10 '18 at 11:26
no! yes !no..........NO! @Rafa Budria – INDIAN May 10 '18 at 11:29
For 3 and 4, the parabola $y=x^2$ cut every characteristic curve ($y=c_1x$) at least once. But as you wish. – Rafa Budría May 10 '18 at 11:39

The differential equation $xu_x + yu_y =2u$ satisfying the initial condition $y = xg(x)$ , $u = f(x)$ with

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