Solving Substitution Steps: Examples Explained

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Discussion Overview

The discussion revolves around the steps involved in solving differential equations using substitution methods, particularly focusing on the implications of defining variables and their derivatives. Participants explore specific examples and the application of the product rule in differentiation.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the validity of the substitution steps involving $$y = ux$$ and the resulting derivative expressions, suggesting a misunderstanding of the differentiation process.
  • Another participant emphasizes the necessity of using the product rule when differentiating $$y = ux$$, indicating that $$\frac{dy}{dx}$$ should not be simplified as proposed.
  • There is a challenge regarding the assumption that $$u$$ is a function of $$x$$, with one participant arguing that restricting $$u$$ to be a constant would limit the generality of the function $$y$$.
  • A later reply suggests that the differential equation should be expressed in terms of $$u = \frac{y}{x}$$ to facilitate substitution, implying that $$u$$ typically varies with $$x$$ unless $$y$$ equals $$x$$.

Areas of Agreement / Disagreement

Participants express differing views on the nature of $$u$$ and its dependence on $$x$$, with no consensus reached on the correct interpretation of the substitution steps or the differentiation process.

Contextual Notes

Participants highlight the importance of understanding the assumptions behind the definitions of $$u$$ and $$y$$, as well as the implications of using the product rule in differentiation. The discussion remains focused on the nuances of these mathematical concepts without resolving the underlying uncertainties.

find_the_fun
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In the context of "solutions by substitutions" in the examples there's a step I don't understand:

If we let $$y = ux$$ then $$dy = u dx + x du$$
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we have $$u = y^{-1}$$ or $$y = u^{-1}$$. We then substitute $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} = -u^{-2} \frac{du}{dx}$$

What is going on in these two examples? If y=ux then [math]\frac{dy}{dx}=u[/math] and [math]dy=u dx \neq udx + xdu[/math]

Same for the other, if [math]y=u^{-1}[/math] then [math]\frac{dy}{dx}=0[/math]?
 
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find_the_fun said:
In the context of "solutions by substitutions" in the examples there's a step I don't understand:
What is going on in these two examples? If y=ux then [math]\frac{dy}{dx}=u[/math] and [math]dy=u dx \neq udx + xdu[/math]

Same for the other, if [math]y=u^{-1}[/math] then [math]\frac{dy}{dx}=0[/math]?

u is a function of x, so to differentiate ux you have to use the product rule, giving (ux)' = u + u'x
 
But how do you know u is a function of x?
 
find_the_fun said:
But how do you know u is a function of x?

Restricting $u$ to be a constant would not allow $y$ to be a general function of $x$.
 
More specifically, your lecturer has explained the steps in a roundabout way. What you really need to do is to get your DE to be a function of (y/x), so that you can substitute u = y/x.

Notice that u will only be a constant if y = x (which it most likely won't).

Thus in general u must be a function of x.
 

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