20070728, 22:19  #1 
Dec 2005
2^{2}·23 Posts 
Differential equation question
Has a deterministic solution to the differential equation y' = y + x been found or is it still unsolved? Also, is there any application where solving that particular differential equation is useful?

20070729, 02:49  #2 
Sep 2002
Vienna, Austria
3×73 Posts 
y(x)=C exp(x)x1

20070729, 09:04  #3 
Dec 2005
5C_{16} Posts 
What is C in the context of an initial value problem?
Last fiddled with by ShiningArcanine on 20070729 at 09:23 Reason: Simplifying question 
20070729, 10:15  #4 
Jun 2007
Moscow,Russia
10000101_{2} Posts 

20070729, 10:41  #5 
Dec 2005
2^{2}×23 Posts 
So is it possible to plot the solution of y as a function of x using this or must it be approximated?
Edit: Also, I am curious, how was the solution that wpolly posted found? Was it found by guessing and proven by substitution? Last fiddled with by ShiningArcanine on 20070729 at 10:56 
20070729, 11:35  #6  
Jun 2007
Moscow,Russia
85_{16} Posts 
Quote:
Generally, there is a family of solutions (infinite numbers): any solution can be plotted by substituing C for some real value. Maybe you have some additional information for your problem: f.e. statement like y(a)=b. If so, you can calculate single C value and then plot the solution,otherwise there are infinite solutions (one solution for each C value) to be plotted. There are many methods for symbolical solving different types of DE. Some of them described at http://eqworld.ipmnet.ru/ru/solutions/ode/odetoc1.htm (russian headers, but english explanation). Your's linear DE can be solved by method,described at http://eqworld.ipmnet.ru/en/solutions/ode/ode0103.pdf You can aslo use (if have) CAS Mathematica or Maple to solve DE. 

20070729, 11:47  #7 
Dec 2005
2^{2}·23 Posts 
Is it known how to solve y(x) for y' = x + y if the vector (0,1) is a solution of y(x)?
Edit: Also, as I asked in my original post, is there an application where solving y' = x + y is useful? Last fiddled with by ShiningArcanine on 20070729 at 11:51 
20070729, 12:15  #8  
Jun 2007
Moscow,Russia
7×19 Posts 
Quote:
If y(0)=1 then (by substituting x with 0 and y with 1) you'll have: 1=C*exp(0)01 or C=2 Thus, you have y(x)=1x+2*exp(x) as a solution which can be plotted. If I still didn't get you, give me the full description of your problem P.S. I don't know the phisical or math application for which this equation can be usefull 

20070729, 12:52  #9 
Dec 2005
134_{8} Posts 
Thanks. That answers my question.

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