Find the solution to the given differential equation

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Homework Help Overview

The discussion revolves around a differential equation involving the relationship between the variables \( z \), \( x \), and \( y \), specifically focusing on the expression \( \left[\dfrac{dz}{dx} - 1 = \dfrac{dy}{dx}\right] \). Participants are seeking clarity on how to manipulate this equation and the implications of differentiating \( z = x + y \) with respect to \( x \).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss differentiating the equation \( z = x + y \) with respect to \( x \) and express confusion regarding the role of \( \dfrac{dz}{dy} \) in the context of the given differential equation. There are inquiries about the implications of the highlighted expression and how to proceed with it.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications about the differentiation process. Some have expressed gratitude for the guidance received, indicating that they are beginning to understand the relationships between the derivatives involved.

Contextual Notes

There is mention of confusion regarding the derivative \( \dfrac{dz}{dy} \) and its relevance to the equation being analyzed. Participants are navigating through the implications of the relationships between the variables without reaching a definitive conclusion.

chwala
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Homework Statement
see attached
Relevant Equations
ode/pde
I need insight on the highlighted in Red on how ##\left[\dfrac{dz}{dx} - 1 = \dfrac{dy}{dx}\right]## otherwise the rest of the steps are clear. I just read that ##\dfrac{dx}{dy} \dfrac{dy}{dz} \dfrac{dz}{dx} =-1##

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chwala said:
I need insight on the highlighted in Red on how ##\left[\dfrac{dz}{dx} - 1 = \dfrac{dy}{dx}\right]## otherwise the rest of the steps are clear.
Just differentiate ##z=x+y## with respect to ##x##.
 
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renormalize said:
Just differentiate ##z=x+y## with respect to ##x##.
I thought of that- the confusion was on what to do with ##\dfrac{dz}{dy}##. Thks.
 
chwala said:
I thought of that- the confusion was on what to do with ##\dfrac{dz}{dy}##. Thks.
The derivative ##dz/dy## doesn't appear in the equation you're trying to derive, so don't worry about it.
 
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renormalize said:
Just differentiate ##z=x+y## with respect to ##x##.
Thanks man!!! i was blind.

##z=x+y##

##\dfrac{dz}{dx} = 1 + \dfrac{dy}{dx}##

##\dfrac{dz}{dx} -1 = \dfrac{dy}{dx}##
 

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