Find the solution to the given differential equation

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SUMMARY

The discussion centers on solving the differential equation involving the relationship between the derivatives of the variables z, x, and y, specifically the equation ##\left[\dfrac{dz}{dx} - 1 = \dfrac{dy}{dx}\right]##. Participants clarify that differentiating the equation ##z=x+y## with respect to x yields ##\dfrac{dz}{dx} = 1 + \dfrac{dy}{dx}##. The confusion regarding the term ##\dfrac{dz}{dy}## is addressed, confirming it is not necessary for this particular derivation. The focus remains on the correct application of differentiation techniques.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically differentiation.
  • Familiarity with the notation of derivatives, such as ##\dfrac{dz}{dx}## and ##\dfrac{dy}{dx}##.
  • Knowledge of implicit differentiation techniques.
  • Ability to manipulate algebraic expressions involving derivatives.
NEXT STEPS
  • Study implicit differentiation methods in calculus.
  • Learn how to apply the chain rule in differentiation.
  • Explore the concept of related rates in calculus.
  • Practice solving differential equations with multiple variables.
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and differential equations, as well as anyone seeking to improve their skills in differentiation and algebraic manipulation.

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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