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

In summary, the attempt at a solution is in the standard 1st order linear ODE form, with P(x) = 1/x and Q(x) = e^(x^2). The u(x) = x and d(uy)/dx = uQ are found, but the R.H.S. is problematic and needs to be integrated multiple times.
  • #1
DryRun
Gold Member
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4
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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  • #2
[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]
 
  • #3
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:
  • #4
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]
 
  • #5
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
 
  • #6
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.
 
  • #7
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?
 

1. What are the steps to solve a 1st order linear ODE?

To solve a 1st order linear ODE, follow these steps:

  • 1. Identify the type of ODE: Is it separable, exact, or linear?
  • 2. Transform the equation into standard form: dy/dx + P(x)y = Q(x)
  • 3. Find the integrating factor, u(x) = e∫P(x)dx
  • 4. Multiply both sides of the equation by u(x)
  • 5. Integrate both sides of the equation and solve for y(x)

2. What is the difference between a separable and exact 1st order linear ODE?

The main difference between a separable and exact 1st order linear ODE is that for a separable ODE, the equation can be written in the form dy/dx = g(x)h(y), where g(x) is a function of x and h(y) is a function of y. This allows for direct integration of both sides. In contrast, an exact ODE requires the use of an integrating factor to make the equation integrable.

3. Can a 1st order linear ODE have a unique solution?

Yes, a 1st order linear ODE can have a unique solution. This is because the general solution to a 1st order linear ODE is given by y(x) = Ce∫P(x)dx + e∫P(x)dx∫Q(x)e-∫P(x)dxdx. The constant C can be adjusted to give a specific solution for a given initial condition, resulting in a unique solution.

4. What is the significance of the integrating factor in solving a 1st order linear ODE?

The integrating factor is a key component in solving a 1st order linear ODE. It is used to transform the equation into an exact form, making it easier to integrate and solve. The integrating factor is found by taking the exponential of the integral of the coefficient of y, and is multiplied to both sides of the equation to make it integrable.

5. Can a 1st order linear ODE have a non-continuous solution?

No, a 1st order linear ODE will always have a continuous solution. This is because the solution is given by integrating both sides of the equation, which results in a continuous function. Any discontinuities in the equation will be reflected in the solution as well.

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