Find the solution to the given differential equation

Click For Summary
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
Gold Member
Messages
2,827
Reaction score
415
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##

1712906561934.png


1712906595606.png
 
Physics news on Phys.org
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##.
 
  • Like
Likes chwala and Orodruin
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.
 
  • Like
  • Wow
Likes docnet and chwala
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}##
 
First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...