Question related to PDE y(z_x)+x(z_y)+z=y

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To solve the PDE:
y(z_x)+x(z_y)+z=y

Use Method of characteristics
a=y
b=x
d-cz=y-z

Thus
dx/y=dy/b=dz/(y-z)

Taking first and second term
xdx=ydy
x^2-y^2=A
x=sqrt(y^2+A)

My question is, at this stage of the calculation, must we account for a negative constant A such that
x=sqrt(y^2-A)
and if not why not?

Thank you
 
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To be honest I'm not familier with the method.

Is it not that your x and y are the independent variables? So what is the domain for the solution function. A is just a constant. It can be positive or negative. I think!
 
Anyone else!?
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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