
Differential Equations Ordinary and Partial Differential Equations Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 27th, 2017, 07:52 AM  #1 
Newbie Joined: Apr 2014 From: Canada Posts: 4 Thanks: 0  Singular solution
How to show that $\displaystyle t=0$ is a solution of $\displaystyle 2t^2e^{x}\,dt+x\sqrt{t}\,dx=0$? I do not know how to calculate $\displaystyle dt$ when $\displaystyle t=0$. Thanks.
Last edited by eulid; November 27th, 2017 at 08:18 AM. 
November 27th, 2017, 09:51 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,874 Thanks: 766 
The way you have phrased this is a little confusing. A first order equation like this can be thought of as an equation describing x as a function of t or as an equation describing t as a function of x. The best I can do in interpreting your question is that you want to show that t, as a function of x, is the constant function t= 0 no matter what x is. In that case, since t is a constant, the derivative is 0: dt/dx= 0 so dt= 0 while dx can be anything. Set t= 0 and dt= 0 and show that the equation is satisfied for all x.

November 27th, 2017, 11:17 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,090 Thanks: 2360 Math Focus: Mainly analysis and algebra 
You don't even need $\mathrm dt$ or $\frac {\mathrm dt}{\mathrm dx}$ for this. $t=0$ is sufficient.


Tags 
singular, solution 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Singular Solution  levanduong123200  Differential Equations  0  August 31st, 2014 12:31 PM 
Singular Value Decomposition  SLUO  Applied Math  1  September 8th, 2013 03:08 AM 
Series solution near regular singular point  triplekite  Calculus  0  October 29th, 2012 03:12 PM 
Series Solution About a Regular Singular Point  mr_dude32  Calculus  1  December 10th, 2008 08:16 PM 
Series Solution About a Regular Singular Point  mr_dude32  Linear Algebra  1  December 10th, 2008 01:15 PM 