Integrating a Problematic Exponential: Solving 1st Order Linear ODE with e^(x^2)

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Homework Help Overview

The discussion revolves around solving a first-order linear ordinary differential equation (ODE) of the form dy/dx + y/x = e^(x^2). Participants are exploring methods to express y in terms of x and an arbitrary constant, while grappling with the complexities introduced by the exponential function.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the standard form of the ODE and the integrating factor method. There are attempts to integrate the right-hand side, particularly focusing on the integral of x.e^(x^2), which some find problematic. Questions arise regarding the nature of the integral and the implications of the exponential function's form.

Discussion Status

Some participants have provided guidance on the integration process and the use of known formulas, while others express confusion about specific aspects, such as the behavior of the integral involving e^(x^2). There is acknowledgment of differing interpretations of the problem, particularly concerning the sign function and its relevance.

Contextual Notes

Participants note that the integral of e^(x^2) is problematic and cannot be expressed in standard terms, which adds to the complexity of the problem. There is also mention of constraints regarding the domain of the solution, specifically for x >= 0.

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Homework Statement
dy/dx + y/x = e^(x^2)
Express y in terms of x and arbitrary constant.

The attempt at a solution
It is in the standard 1st order linear ODE form.

P(x) = 1/x
Q(x) = e^(x^2)

u(x) = x (after calculation)

So, d(uy)/dx = uQ

d(uy)/dx = x.e^(x^2)
I have to integrate both sides w.r.t.x

Finding the R.H.S is problematic though. As it seems infinite, from what I've understood from my calculations.

Integral of x.e^(x^2) (done by partial integration)
Let U = x, so dU/dx = 1
Let dV = e^(x^2), so V = [e^(x^2)]/2x (is this correct?)

xy = x.[e^(x^2)]/2x - integral of [e^(x^2)]/2x.(1).dx

Then i have to integrate the R.H.S again and again and again, as i can't get rid of e^(x^2) with a multiple of x always in the denominator.

Any advice?
 
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[tex]y'+a(x)y = b(x)[/tex]
[tex]y=e^{-\int a(x)dx} \int e^{\int a(x)dx}b(x)dx[/tex]
 
Hi Quinzio

I know about that formula. I used it in the first place, and the spot where I'm stuck is due to b(x) in your equation being e^(x^2). The power of e contains x which is giving me quite some trouble with this problem.
 
Last edited:
Ok, but you got an x which is its derivative (of [itex]x^2[/itex]).
It goes:
[tex]a(x)=1/x[/tex]
[tex]\int a(x)dx=ln|x|[/tex]
[tex]e^{\int a(x)dx}=|x|[/tex]
[tex]\int e^{\int a(x)dx}b(x)dx=|x|e^{x^2}=\frac{1}{2}e^{x^2}*sgn(x)[/tex]
 
I did some research and it appears that the integral of e^(x^2) is infinity and cannot be expressed in normal terms.

However, it is curious that my calculus notes give this as the answer to this problem:

y= [e^(x^2)]/2x + A/x
 
Your notes show the right answer.
I just wrote a part of it, what I thought was the harder.
The answer in your notes is for x>=0 only, anyway.
 
This is the last line where I've been able to reach:

xy = x.[e^(x^2)]/2x - integral of [e^(x^2)]/2x.(1).dx

I'm stuck at the R.H.S. where i need to integrate [e^(x^2)]/2x

I realize that i know nothing about the sign function. If that is a requirement to solve this problem, then i won't be able to do this.

Is there any easy guide that you could possibly recommend to learn about the sign function?
 

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