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 2007-07-28, 22:19 #1 ShiningArcanine     Dec 2005 22·23 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?
 2007-07-29, 02:49 #2 wpolly     Sep 2002 Vienna, Austria 3×73 Posts y(x)=C exp(x)-x-1
 2007-07-29, 09:04 #3 ShiningArcanine     Dec 2005 9210 Posts What is C in the context of an initial value problem? Last fiddled with by ShiningArcanine on 2007-07-29 at 09:23 Reason: Simplifying question
2007-07-29, 10:15   #4
VolMike

Jun 2007
Moscow,Russia

8516 Posts

Quote:
 Originally Posted by ShiningArcanine What is C in the context of an initial value problem?
C is some complex (in general case) value

 2007-07-29, 10:41 #5 ShiningArcanine     Dec 2005 22×23 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 2007-07-29 at 10:56
2007-07-29, 11:35   #6
VolMike

Jun 2007
Moscow,Russia

7·19 Posts

Quote:
 Originally Posted by ShiningArcanine 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?

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/ode-toc1.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.

 2007-07-29, 11:47 #7 ShiningArcanine     Dec 2005 9210 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 2007-07-29 at 11:51
2007-07-29, 12:15   #8
VolMike

Jun 2007
Moscow,Russia

7·19 Posts

Quote:
 Originally Posted by ShiningArcanine 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?
It seems I undrestand you; vector solution (0,1) of y(x) is the same with y(0)=1. If so, you just need to solve linear equation respectively to C. I.e. you have common solution y(x)=C* exp(x)-x-1 of y'=x+y (it must be found before, by CAS or manually)

If y(0)=1 then (by substituting x with 0 and y with 1) you'll have:
1=C*exp(0)-0-1 or C=2
Thus, you have y(x)=-1-x+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

 2007-07-29, 12:52 #9 ShiningArcanine     Dec 2005 22·23 Posts Thanks. That answers my question.

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