Difficult first order linear differential equation

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SUMMARY

The discussion focuses on solving the first-order linear differential equation given by $y'=\dfrac{x^3-y^3}{x-y}$. The Wolfram Development Platform, which is based on Mathematica, provides a complex solution involving Hermite $H_n(x)$ functions and the hypergeometric $_1F_1$ function. An alternative solution presented simplifies to the form $y'=x^2 + x y + y^2$, indicating a more manageable approach to the problem.

PREREQUISITES
  • Understanding of first-order linear differential equations
  • Familiarity with Hermite functions and hypergeometric functions
  • Experience with the Wolfram Development Platform (Mathematica)
  • Basic knowledge of differential equation solvers
NEXT STEPS
  • Explore the properties and applications of Hermite functions
  • Learn about hypergeometric functions and their role in differential equations
  • Investigate alternative methods for solving first-order linear differential equations
  • Practice using the Wolfram Development Platform for solving complex differential equations
USEFUL FOR

Mathematics students, researchers in applied mathematics, and professionals dealing with differential equations will benefit from this discussion.

WMDhamnekar
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Hello, I want to solve the following differential equation. $y'=\dfrac{x^3-y^3}{x-y}$. How to solve it?
 
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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.

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

y'=x^2 + x y + y^2
 

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