Solving Differential Equation: Substitution Method

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SUMMARY

The discussion focuses on solving a differential equation using the substitution method, specifically with the substitutions \( x = U + \frac{3}{2} \) and \( y = W - \frac{1}{2} \). The user encountered difficulties in solving for \( z \) after integrating, leading to an implicit solution involving \( \arctan z - \frac{1}{2} \ln(z^2 + 1) = \ln(U) + C \). Another participant confirmed that this implicit solution is valid and suggested that the integration steps should be verified for accuracy.

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Homework Statement


Solve: http://img384.imageshack.us/img384/60/clipboard01az7.jpg

Homework Equations


DE stuff

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 ,[/URL] but http://img167.imageshack.us/img167/8297/clipboard03db9.jpg ,[/URL] so http://img175.imageshack.us/img175/3223/clipboard04xt2.jpg .[/URL]

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

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

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

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

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

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

http://img521.imageshack.us/img521/3638/clipboard13nj5.jpg .[/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
 
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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.
 
Surely there is a way to solve it explicitely
 
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.)