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

## Summary:

How to solve ##f'(t) = -t + t^3f(t)##
• Delta2

etotheipi
Gold Member
2020 Award
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

Last edited:
• Delta2
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.

• Delta2
fresh_42
Mentor
• Delta2
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.

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