MHB Difficult first order linear differential equation

WMDhamnekar
MHB
Messages
376
Reaction score
28
Hello, I want to solve the following differential equation. $y'=\dfrac{x^3-y^3}{x-y}$. How to solve it?
 
Physics news on Phys.org
Wolfram Development Platform (essentially Mathematica) gives a horrendous answer: extremely complicated with Hermite $H_n(x)$ functions and the hypergeometric $_1F_1$ function. In what context did this problem come up, and what have you tried?
 
Ackbach said:
Wolfram Development Platform (essentially Mathematica) gives a horrendous answer: extremely complicated with Hermite $H_n(x)$ functions and the hypergeometric $_1F_1$ function. In what context did this problem come up, and what have you tried?

I got the following answer from my differential equation solver.

\[y(x)=\frac{x^3}{6}(C_1(C_1+1)+3C_1+2)+\frac{x^5}{120}(C^3_1+C_1(74C_1+9)+31C_1+8)+C_1+C_1x+\frac{3C_1x^2}{2}+\frac{C_1x^4}{12}(C_1+4)+\mathcal{O}\]
 
Last edited by a moderator:
You can reduce the expression to this

y'=x^2 + x y + y^2
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
Back
Top