# Homework Help: DE question

1. Nov 24, 2008

### -Vitaly-

1. The problem statement, all variables and given/known data
Solve: http://img384.imageshack.us/img384/60/clipboard01az7.jpg [Broken]

2. Relevant equations
DE stuff

3. The attempt at a solution
I started by doing substitution: x=U+3/2 and y=W-1/2

So, it gave me http://img522.imageshack.us/img522/3172/clipboard02ly8.jpg [Broken],[/URL] but http://img167.imageshack.us/img167/8297/clipboard03db9.jpg [Broken],[/URL] so http://img175.imageshack.us/img175/3223/clipboard04xt2.jpg [Broken].[/URL]

Therefore, http://img237.imageshack.us/img237/2264/clipboard05iu8.jpg [Broken]

And, http://img377.imageshack.us/img377/615/clipboard06ez2.jpg [Broken].[/URL] Substitution z=W/U. Giving, http://img139.imageshack.us/img139/5639/clipboard07hb7.jpg [Broken] (*).

And, http://img399.imageshack.us/img399/2373/clipboard08qj2.jpg [Broken]

Rearranging gives http://img114.imageshack.us/img114/1830/clipboard09tm2.jpg [Broken].[/URL] Substitute in (*)

http://img140.imageshack.us/img140/8541/clipboard10bz2.jpg [Broken] => http://img140.imageshack.us/img140/6706/clipboard11ao0.jpg [Broken] => http://img135.imageshack.us/img135/8163/clipboard12ee3.jpg [Broken]

Dividing 1/(1+z2) and -z/(1+z2) and integrating gives:

http://img521.imageshack.us/img521/3638/clipboard13nj5.jpg [Broken].[/URL] But now I'm stuck :( because I can't solve this for z and go back to y and x :( maybe I made an arithmetical mistake or maybe there is a better method?
Thank you

Last edited by a moderator: May 3, 2017
2. Nov 24, 2008

### µ³

I don't know how correct the method you detailed is but a lot of times for a differential equation, an implicit solution is sufficient.

3. Nov 25, 2008

### -Vitaly-

Surely there is a way to solve it explicitely

4. Nov 25, 2008

### HallsofIvy

Why "surely". You don't need to solve
$$arctan z-\frac{1}{2}ln(z^2+ 1)= ln(U)+ C$$

z= W/U so that is
$$arctan W/U-\frac{1}{2}ln(\frac{W^2}{U^2}+ 1)= ln(U)+ C$$
and U= x- 3/2, W= y+ 1/2 so
$$arctan\frac{y+ 1/2}{x-3/2}- \frac{1}{2}ln(\frac{(y+1/2)^2}{(x-3/2)^2}= ln(x- 3/2)+ C$$
That's a perfectly good solution. (Assuming your integration was correct.)