# Solve 1st order linear ODE

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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.

Related Calculus and Beyond Homework Help News on Phys.org
$$y'+a(x)y = b(x)$$
$$y=e^{-\int a(x)dx} \int e^{\int a(x)dx}b(x)dx$$

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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 $x^2$).
It goes:
$$a(x)=1/x$$
$$\int a(x)dx=ln|x|$$
$$e^{\int a(x)dx}=|x|$$
$$\int e^{\int a(x)dx}b(x)dx=|x|e^{x^2}=\frac{1}{2}e^{x^2}*sgn(x)$$

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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

I just wrote a part of it, what I thought was the harder.

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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?