Differential equation (to solve analytically)

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SUMMARY

The discussion focuses on solving the differential equation (x^2+1)y' = x^2+x-1+4xy analytically. The equation can be rewritten as a linear differential equation in the form (x^2+1)dy/dx - 4xy = x^2+x-1. An integrating factor, e^{-∫(4x/(x^2+1))dx}, is essential for transforming the left side into an exact derivative, facilitating the solution process. Participants emphasize the importance of correctly applying the integrating factor to solve the equation accurately.

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  • Understanding of linear differential equations
  • Familiarity with integrating factors in differential equations
  • Knowledge of basic calculus, specifically integration techniques
  • Ability to manipulate algebraic expressions involving derivatives
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  • Study the method of integrating factors for linear differential equations
  • Learn about exact differential equations and their solutions
  • Explore advanced integration techniques, particularly for rational functions
  • Practice solving various forms of differential equations analytically
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Students studying differential equations, educators teaching calculus, and anyone interested in analytical methods for solving mathematical equations.

dragonblood
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(x^2+1)y'=x^2+x-1+4xy
How can I solve this equation analytically?
I have almost no idea. I thought that y might be a series...please help :)

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If you write that as (x^2+ 1)dy/dx- 4xy= x^2+ x- 1 or
\frac{dy}{dx}- \frac{4x}{x^2+ 1}y= \frac{x^2+ x- 1}{x^2+ 1}[/itex]<br /> a linear differential equation. Then <br /> e^{\int {4x}{x^2+1} dx} is an integrating factor. Multiplying the entire equation by it will make the left side an &quot;exact&quot; derivative.
 
Let me just clarify that integrating factor for you ivy =] e^{- \int \frac{4x}{x^2+1} dx}
 
Allright :) Thanks peeps!
 
Dang! Dropped a sign, didn't I?
 

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