How to solve f'(t) = -t + t^3.f(t)

[murshid_islam]

TL;DR

How to solve $f'(t) = -t + t^3f(t)$

I came across that differential equation while trying to compute something else. Here's a link to that: https://www.physicsforums.com/threa...fferentiating-under-the-integral-sign.998249/

And since I have studied differential equations more than 15 years ago, I can't remember how to (or if there's a way to) solve this:
$f'(t) = -t + t^3f(t)$

Any ideas?

 

[etotheipi]

I guess you could try an integrating factor $\mu = \text{exp} \left(- \frac{1}{4} t^4 \right)$, in which case the solution is$$f(t) = \frac{\int -t \, \text{exp} \left(- \frac{1}{4} t^4 \right)}{ \text{exp} \left(- \frac{1}{4} t^4 \right)}$$but now you need to do that integral

 

[murshid_islam]

I guess you could try an integrating factor $\mu = \text{exp} \left(- \frac{1}{4} t^4 \right)$, in which case the solution is$$f(t) = \frac{\int -t \, \text{exp} \left(- \frac{1}{4} t^4 \right)}{ \text{exp} \left(- \frac{1}{4} t^4 \right)}$$but now you need to do that integral and I don't think it's one you can do analytically without special functions. So, maybe there is a better approach

After the substitution $u = \frac{t^2}{2}$, the integral becomes $\int e^{-u^2}du$, which brings me back to square one because that's the integral I was trying to compute here: Gaussian integral by differentiating under the integral sign, which led to that differential equation in the first place.

 

[fresh_42]

Here is how the Gaussian integral is done:
https://en.wikipedia.org/wiki/Gaussian_integral
Compute $[\int \exp(ax^2)dx]^2$, switch to polar coordinates and calculate the result.

 

[murshid_islam]

Here is how the Gaussian integral is done:
https://en.wikipedia.org/wiki/Gaussian_integral
Compute $[\int \exp(ax^2)dx]^2$, switch to polar coordinates and calculate the result.

Yeah, I do know the standard method of computing it by squaring it and changing to polar coordinates, but I wanted to do it with this method of differentiating under the integral sign that I recently learned.

 

[HomogenousCow]

The integral is non-elementary, there is no way to compute the indefinite integral in terms of elementary functions.