First order linear differential Equation

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Discussion Overview

The discussion revolves around solving a first-order linear differential equation of the form $$(x^2+1)y'-2xy=x^2+1$$ with the initial condition $$y(1)=\frac{\pi}{2}$$. Participants explore the steps involved in simplifying the equation and applying logarithmic and exponential rules.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the differential equation and attempts to simplify it by dividing through by $$x^2+1$$, leading to a new form of the equation.
  • Another participant suggests simplifying the expression further and discusses the relationship between logarithms and exponentials.
  • A participant questions their understanding of the simplification, proposing that $$e^{-ln(x^2+1)}$$ equals $$-x^2-1$$.
  • Participants clarify the rules of logarithms, emphasizing that $$e^{-ln(x^2+1)}$$ should be interpreted correctly, leading to the conclusion that it equals $$\frac{1}{x^2+1}$$.

Areas of Agreement / Disagreement

There is no consensus on the initial simplification steps, as participants express uncertainty and correct each other regarding the application of logarithmic and exponential rules.

Contextual Notes

Participants rely on specific logarithmic properties and the correct interpretation of exponential functions, which may depend on their understanding of these mathematical concepts.

Who May Find This Useful

Students and individuals interested in differential equations, particularly those seeking clarification on logarithmic and exponential relationships in mathematical expressions.

Petrus
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Hello MHB,
$$(x^2+1)y'-2xy=x^2+1$$ if $$y(1)=\frac{\pi}{2}$$What I have done:
Divide evrything by $$x^2+1$$ and we got
$$y'-\frac{2xy}{x^2+1}=1$$
we got the integer factor as $$e^{^-\int\frac{2x}{x^2+1}}= e^{-ln(x^2+1)}$$
Now I get
$$(e^{-ln(x^2+1)}y)'=e^{-ln(x^2+1)}$$
and this lead me to something wrong, I am doing something wrong or?

Regards,
$$|\pi\rangle$$
 
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Try simplifying the expression first , remember logarithm and exponential are (inverse) ...
 
ZaidAlyafey said:
Try simplifying the expression first , remember logarithm and exponential are (inverse) ...
that means if I got right
$$e^{-ln(x^2+1)}=-x^2-1$$?

Regards,
$$|\pi\rangle$$
 
No , remember the rules of logarithm $$\log x^n = n\log x $$ .
 
ZaidAlyafey said:
No , remember the rules of logarithm $$\log x^n = n\log x $$ .
Ahhh I see now! $$e^{-ln(x^2+1)}=e^{ln((x^2+1)^{-1})}= \frac{1}{x^2+1}$$
Thanks a lot Zaid! I should be able to continue!:) Thanks for fast responed and taking your time!

Regards,
$$|\pi\rangle$$
 

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