Find the solution to the given differential equation

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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##

Printed problem: use the substitution z = x + y to solve dy/dx = (1 + 3x + 3y)/(3x + 3y - 1) for y = 0 when x = 1, giving the answer as a ln(x+y) + b(x-y) + c = 0 with constants a, b, c, marked [7].


Printed marking-scheme table of solution steps for a differential equation, with mark codes B1, M1, A1; the first row shows dy/dx = dz/dx - 1, with a red mark drawn over it.
 
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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}##