Find the solution to the given differential equation

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The discussion focuses on solving the differential equation related to the expression ##\left[\dfrac{dz}{dx} - 1 = \dfrac{dy}{dx}\right]##. Participants clarify that differentiating the equation ##z=x+y## with respect to ##x## leads to the necessary insights. The confusion regarding the term ##\dfrac{dz}{dy}## is addressed, emphasizing that it is not relevant to the equation being derived. Ultimately, the correct differentiation yields ##\dfrac{dz}{dx} = 1 + \dfrac{dy}{dx}##, confirming the relationship needed for the solution. This exchange highlights the importance of focusing on relevant derivatives in differential equations.
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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