My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Thanks Tree1Thanks
  • 1 Post By v8archie
Reply
 
LinkBack Thread Tools Display Modes
November 25th, 2017, 11:58 PM   #1
Newbie
 
Joined: Apr 2014
From: Canada

Posts: 4
Thanks: 0

general solution

$y=0, x<0$ and $y=0, x>0$ are solutions of $y'=\frac{y^2}{x^2+xy}$. The solutions is not defined at $x=0$ because $x=0$ is impossible for the differential equation $y'=\frac{y^2}{x^2+xy}$. However, substituting $c=0$ into the general solution $y=ce^{-y/x}$, we get $y=0$ not $y=0, x<0$ and not $y=0, x>0$. How to modify the general solution $y=ce^{-y/x}$ so that the particular solution for $c=0$ gives the two solutions $y=0, x<0$ and $y=0, x>0$?
eulid is offline  
 
November 26th, 2017, 04:44 AM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,404
Thanks: 2477

Math Focus: Mainly analysis and algebra
$$y=ce^{-\frac y x} \quad (x \ne 0)$$
Or, even better
$$y=\begin{cases}c_1e^{-\frac y x} & (x \lt 0) \\ c_2e^{-\frac y x} & (x \gt 0) \end{cases}$$
Notice that you now need two sets of initial conditions to fix a unique solution.

Another way would be to multiply the original equation by $x$ so that $y=0$ becomes a valid solution for all $x$.
Thanks from eulid

Last edited by v8archie; November 26th, 2017 at 04:47 AM.
v8archie is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
general, solution



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Getting the General solution The_Ys_Guy Differential Equations 10 April 19th, 2018 12:36 PM
General Solution BonaviaFx Trigonometry 1 November 19th, 2015 01:16 PM
General solution of ODE Tooperoo Calculus 4 October 16th, 2013 01:10 AM
General Solution AzraaBux Algebra 4 May 31st, 2013 07:13 AM
General Solution of a PDE mathbalarka Calculus 11 May 5th, 2013 11:01 AM





Copyright © 2018 My Math Forum. All rights reserved.