My Math Forum  

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

Differential Equations Ordinary and Partial Differential Equations Math Forum


Thanks Tree3Thanks
  • 1 Post By v8archie
  • 1 Post By skipjack
  • 1 Post By v8archie
Reply
 
LinkBack Thread Tools Display Modes
November 7th, 2017, 06:49 AM   #1
Newbie
 
Joined: Apr 2016
From: Wonderland

Posts: 14
Thanks: 0

How to tell when y = 0 is a solution?

Hi there, when solving differential equations, how do you know when y = 0 is a solution?

For example:
Solve the differential equation y - (x-2) dy/dx = 0

I don't really understand the how "y = 0" can be a solution. Any guides? And is this applicable for first order & second order (non) homogeneous equations?
Appletree is offline  
 
November 7th, 2017, 07:32 AM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,034
Thanks: 2342

Math Focus: Mainly analysis and algebra
If $y=0$ is a solution, it is constant. So $\frac{\mathrm dy}{\mathrm dx}=0$.

Plugging into the given equation, we see that it is satisfied for all values of $x$ ($0 - (x-2)0 = 0$), so $y=0$ is indeed a solution.
Thanks from topsquark
v8archie is online now  
November 7th, 2017, 08:46 AM   #3
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,034
Thanks: 2342

Math Focus: Mainly analysis and algebra
This approach works for all equations because all we are doing is checking whether or not a given solution satisfies the given equation.
v8archie is online now  
November 7th, 2017, 09:36 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 18,155
Thanks: 1422

This overlooks why one would be interested in testing y = 0 in the first place.

I suggest taking a further look at the slightly different ODE, 2y - (x-2) dy/dx = 0, where x and y are real.
Thanks from topsquark
skipjack is offline  
November 7th, 2017, 11:23 AM   #5
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,034
Thanks: 2342

Math Focus: Mainly analysis and algebra
I'm not sure what you are getting at there Skipjack. The two equations can be solved in exactly the same ways.

\begin{align*}
2y - (x-2)y' &= 0 \\
y' + \frac{2}{2-x} y &= 0 &&& \int \frac{2}{2-x} \,\mathrm dx & = -2\ln{(2-x)} = \ln{\frac1{(x-2)^2}} \\
&&&& p(x) &= e^{\ln{\frac1{(x-2)^2}}} = \frac1{(x-2)^2} \\
\frac1{(x-2)^2}y' - \frac2{(x-2)^3}y &= 0 \\
\frac1{(x-2)^2} y &= c \\
y &= c(x-2)^2
\end{align*}
which gives $y=0$ when $c=0$.

Alternatively, the both equations can be solved by separating variables.
\begin{align*}
y - (x-2)y' &= 0 \\
\int \frac1y \frac{\mathrm dy}{\mathrm dx}\, \mathrm dx &= \int \frac1{x-2} \,\mathrm dx \\
\ln |y| &= \ln{A|x-2|} & (A \gt 0) \\
y &= \pm A(x-2) & (A \gt 0) \\
y &= c(x-2) & (c \ne 0)
\end{align*}
And in this case we don't get $y=0$ because $c=0$ is not permitted (because the logarithm of zero is not defined).

We lost the solution $y=0$ at line 2 where we divided by $y$ because this is only possible when $y \ne 0$. Thus we manually verify whether or not $y=0$ is a solution as above.

In the equation $(y-2) - (x-2)y' = 0$ (solving by separating variables) we divide by $(y-2)$ and so we'd have to verify $y-2=0 \implies y=2$ manually.
Thanks from topsquark

Last edited by v8archie; November 7th, 2017 at 11:46 AM.
v8archie is online now  
November 7th, 2017, 11:34 AM   #6
Senior Member
 
Joined: Jun 2015
From: England

Posts: 697
Thanks: 199

Archie, I would be very grateful to learn how you managed to lay that math out on the page like you did.

studiot is offline  
November 7th, 2017, 11:46 AM   #7
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,034
Thanks: 2342

Math Focus: Mainly analysis and algebra
Quote the post to see the $\LaTeX$.
v8archie is online now  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
solution



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
x^x=2 solution? ricsi046 Calculus 4 December 15th, 2014 01:55 PM
Why my solution is different from sample solution? rain Real Analysis 1 July 17th, 2013 01:28 PM
only 1 solution alexmath Calculus 2 May 24th, 2013 06:26 PM
my correct solution but another solution? davedave Calculus 1 January 31st, 2012 03:48 PM
trying to make a solution (liquid solution...) problem... TreeTruffle Algebra 2 March 27th, 2010 02:22 AM





Copyright © 2017 My Math Forum. All rights reserved.