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