# Help nedded to solve a D.E. using substitution

1. Nov 16, 2008

### bobmerhebi

1. The problem statement, all variables and given/known data

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

2. Relevant equations

y = ux

3. The attempt at a solution

y = ux implies dy = udx + xdu

so -xudx + (x + x$$\sqrt{}u$$)(udx + xdu) =0

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

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

hence dx/x + du/(u$$\sqrt{}u$$) + du/u = 0

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

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

ln x + ln y - lnx - 2$$\sqrt{}x$$/$$\sqrt{}y$$ = c1

so ln y - 2$$\sqrt{}x$$/$$\sqrt{}y$$ = 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 plz. my process is right isn't it? how should i continue?

2. Nov 16, 2008

### HallsofIvy

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

3. Nov 16, 2008

### bobmerhebi

thank u Sir for the help.

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

thx again