Differential equation question

[ShiningArcanine]

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?

[wpolly]

y(x)=C exp(x)-x-1

[ShiningArcanine]

What is C in the context of an initial value problem?

[VolMike]

Originally Posted by ShiningArcanine

What is C in the context of an initial value problem?

C is some complex (in general case) value

[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?

[VolMike]

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.

[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?

[VolMike]

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

[ShiningArcanine]

Thanks. That answers my question.