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Partial Differential Equation with square roots

  • #1
<Moderator's note: Moved from a technical forum and thus no template.>

Hi everyone,
I have encountered a partial differential equation with square roots which I don't have a clue in solving it. After letting z=F(x)+G(y), I can't really figure out the next step. I tried squaring both sides but the square root still exists. Any help would be appreciated, thank you!

upload_2018-6-5_4-7-12.png
 

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  • #2
BvU
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Hello Johnson, :welcome:

Please post in a homework forum and use the template. See Guidelines . We need an attempt at solution (posted) to be allowed to help out.

In the mean time:
How would you solve ##\sqrt{\partial z\over \partial x} = x ## ?
 
  • #3
Thank you for your reply, my attempt to your problem are as follow:
\begin{aligned}\dfrac {\partial z}{\partial x}=x^{2}\\ \partial z=x^{2}\partial x\\ z=\dfrac {x^{3}}{3}+c\end{aligned}
My attempt to my problem are as follow:
I try to reproduce the equation to a 1st order differential equation,
Let z(x,y)= F(x)+G(y)
Therefore the equation becomes (F'(x))^1/2+(G'(y))^1/2=x
Taking square on both sides still doesn't eliminate the the square root and squaring it one more time involves F'(x)^2.
 
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  • #4
BvU
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my attempt to your problem
Good. Notice anything remarkable in relation to your problem ?
 
  • #5
Sorry I don't quite understand the relation, but my attempt on continuing the question with your hint:
\begin{aligned}\sqrt {\dfrac {\partial z}{\partial x}}+\sqrt {\dfrac {\partial z}{\partial y}}=x\\
\sqrt {\dfrac {\partial z}{\partial x}}=x\ & \sqrt {\dfrac {\partial z}{\partial y}}=0\\
\therefore \dfrac {\partial z}{\partial x}=x^{2},\dfrac {\partial z}{\partial y}=0\\
z=\dfrac {x^{3}}{3}+G\left( y\right) =\dfrac {x^{3}}{3}+C\end{aligned}
 
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  • #6
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That's what I meant: the solution to my problem is also a solution for your problem :smile:
I must say that the separation of variables thing is useful but a bit confusing: you were supposed to conclude that if ##z = F(x) + G(y)## then ##\sqrt{\partial z\over \partial y} ## is a function of ##y## and the right hand side of the equation is only a function of x, square root or no square root on the left. In other words, satisfied if ##G(y) = 0##.
 
  • #7
That's what I meant: the solution to my problem is also a solution for your problem :smile:
I must say that the separation of variables thing is useful but a bit confusing: you were supposed to conclude that if ##z = F(x) + G(y)## then ##\sqrt{\partial z\over \partial y} ## is a function of ##y## and the right hand side of the equation is only a function of x, square root or no square root on the left. In other words, satisfied if ##G(y) = 0##.
Thanks! Helped me out a lot.
 

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