Help nedded to solve a D.E. using substitution

[bobmerhebi]

Homework Statement

Use the appropriate substitution to solve the following D.E.: -ydx + (x + [tex]\sqrt{}xy[/tex])dy = 0

Homework Equations

y = ux

The Attempt at a Solution

y = ux implies dy = udx + xdu

so -xudx + (x + x[tex]\sqrt{}u[/tex])(udx + xdu) =0

we then get after some simplificaion: xu[tex]\sqrt{}u[/tex] dx + x² (1 + [tex]\sqrt{}u[/tex])du = 0

so (1/x).u[tex]\sqrt{}u[/tex]dx + (1 + [tex]\sqrt{}u[/tex])du = 0

hence dx/x + du/(u[tex]\sqrt{}u[/tex]) + du/u = 0

now we have after integrating: lnx + lnu - 2/[tex]\sqrt{}u[/tex] = c₁

substituting bk u= y/x we have: ln x + ln (y/x) - 2[tex]\sqrt{}x[/tex]/[tex]\sqrt{}y[/tex] = c₁

ln x + ln y - lnx - 2[tex]\sqrt{}x[/tex]/[tex]\sqrt{}y[/tex] = c₁

so ln y - 2[tex]\sqrt{}x[/tex]/[tex]\sqrt{}y[/tex] = c₁

here i got stuck. i couldn;t continue although i know that the answer should be : 4x = y(ln|y| - c)²

need help in this please. my process is right isn't it? how should i continue?

 

[HallsofIvy]

The rest is just basic algebra!
From [itex]ln|y| 2\sqrt{x}/\sqrt{y}= c1[/itex], [itex]2\sqrt{x}/sqrt{y}= ln|-|c1[/itex],
[itex]2\sqrt{x}= (ln|y|- c1)\sqrt{y}[/itex]. Squaring [itex]4x= (ln y- c1)^2y[/itex].

 

[bobmerhebi]

thank u Sir for the help.

I feel like a dum after it appeared to be that apparent.

thx again