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 Differential Equations Ordinary and Partial Differential Equations Math Forum

November 27th, 2016, 09:19 AM   #1
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help separate eq

Please help; I don't know how to start separating the equation.
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Last edited by skipjack; November 27th, 2016 at 06:37 PM. November 27th, 2016, 10:47 AM   #2
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Quote:
 Originally Posted by ibanez1608 please help i dont know how to start separating the eq
$\displaystyle xyy' = (x + y)^2$ for y(1) = 0.

You might be able to separate this (you'll need a change of variables if you can even do it) but note that your initial condition isn't part of the solution set:
$\displaystyle xyy' = 1 \cdot 0 \cdot y'(1) = 0$ for any value of y'(1)

but
$\displaystyle (x + y)^2 = (1 + 0)^2 = 1$

-Dan November 27th, 2016, 03:18 PM #3 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 \displaystyle \begin{align*} x\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} &= \left( x + y \right) ^2 \\ &= x^2 + 2\,x\,y + y^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{x}{y} + 2 + \frac{y}{x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\frac{y}{x}} + 2 + \frac{y}{x} \end{align*} Now let \displaystyle \begin{align*} u = \frac{y}{x} \implies y = u\,x \end{align*}, then \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} \end{align*} giving \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\frac{y}{x}} + 2 + \frac{y}{x} \\ u + x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{u} + 2 + u \\ x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{u} + 2 \\ x\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1 + 2\,u}{u} \\ \frac{u}{1 + 2\,u}\,\frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{1}{x} \\ \int{ \frac{u}{1 + 2\,u}\,\frac{\mathrm{d}u}{\mathrm{d}x} \,\mathrm{d}x} &= \int{\frac{1}{x}\,\mathrm{d}x} \\ \int{ \frac{u}{1 + 2\,u} \,\mathrm{d}u} &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{2} \int{ \frac{2\,u}{1 + 2\,u} \,\mathrm{d}u } &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{2} \int{ \left( 1 - \frac{1}{1 + 2\,u} \right) \,\mathrm{d}u } &= \ln{ \left| x \right| } +C_1 \\ \frac{1}{4} \int{ \left( 2 - \frac{2}{1 + 2\,u} \right) \,\mathrm{d}u } &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{4} \, \left( 2\,u - \ln{ \left| 1 + 2\,u \right| } \right) + C_2 &= \ln{ \left| x \right| } + C_1 \\ \frac{1}{2}\,u - \frac{1}{4}\,\ln{ \left| 1 + 2\,u \right| } &= \ln{ \left| x \right| } + C \textrm{ where } C = C_1 - C_2 \\ \frac{y}{2\,x} - \frac{1}{4}\, \ln{ \left| 1 + \frac{2\,y}{x} \right| } &= \ln{ \left| x \right| } + C \end{align*} Now since \displaystyle \begin{align*} y \left( 1 \right) = 0 \end{align*} that means \displaystyle \begin{align*} \frac{0}{2\cdot 1} - \frac{1}{4} \,\ln{ \left| 1 + \frac{2\cdot 0}{1} \right| } &= \ln{ \left| 1 \right| } + C \\ 0 - \frac{1}{4}\, \ln{\left( 1 \right) } &= \ln{ \left( 1 \right) } + C \\ C &= 0 \end{align*} and thus \displaystyle \begin{align*} \frac{y}{2\,x} - \frac{1}{4}\, \ln{ \left| 1 + \frac{2\,y}{x} \right| } &= \ln{ \left| x \right| } \end{align*}. Thanks from topsquark and ibanez1608 Last edited by skipjack; November 27th, 2016 at 05:00 PM. November 27th, 2016, 03:31 PM   #4
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Quote:
 Originally Posted by Prove It Now since \displaystyle \begin{align*} y \left( 1 \right) = 0 \end{align*} that means \displaystyle \begin{align*} \frac{0}{2\cdot 1} - \frac{1}{4} \,\ln{ \left| 1 + \frac{2\cdot 0}{1} \right| } &= \ln{ \left| 1 \right| } + C \\ 0 - \frac{1}{4}\, \ln{\left( 1 \right) } &= \ln{ \left( 1 \right) } + C \\ C &= 0 \end{align*} and thus \displaystyle \begin{align*} \frac{y}{2\,x} - \frac{1}{4}\, \ln{ \left| 1 + \frac{2\,y}{x} \right| } &= \ln{ \left| x \right| } \end{align*}.
Nicely done. But I still object to the initial condition. In order to find the solution to the DEq you divided both sides by xy. So y(1) = 0 is still outside the solution set.

-Dan November 27th, 2016, 08:00 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 The graph for the equation obtained contains the point (1, 0), but doesn't have a slope there, so the differential equation isn't satisfied there. Hence there is no solution. November 28th, 2016, 04:35 AM #6 Newbie   Joined: Nov 2016 From: israel Posts: 5 Thanks: 0 ty prove it Tags separate Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ricsi046 Number Theory 3 September 4th, 2014 07:18 PM sarah1994 Algebra 2 October 29th, 2013 12:16 PM arslan894 Elementary Math 5 February 22nd, 2012 01:23 PM mbradar2 Calculus 1 December 2nd, 2010 09:03 AM gplush Algebra 1 August 25th, 2009 11:17 AM

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