Seperable Partial Differential Equation

  • Thread starter CINA
  • Start date
  • #1
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Homework Statement



[tex]u_{x}=u_{y}+u[/tex]

Homework Equations



Separation of variables

The Attempt at a Solution



It reduces to [tex]\frac{X'}{X}=\frac{Y'}{Y}+1[/tex]

My question is does the 1 change at all how I use [tex]\lambda^{2}[/tex]? In other words, will my solution still break down into this:

Case 1: [tex]\lambda^{2}=0[/tex]

[tex]\frac{X'}{X}=0 [/tex] and [tex]\frac{Y'}{Y}+1=0[/tex]

Case 2: [tex]-\lambda^{2}[/tex]

[tex]\frac{X'}{X}=-\lambda^{2} [/tex] and [tex]\frac{Y'}{Y}+1=-\lambda^{2}[/tex]

Case 3: [tex]\lambda^{2}[/tex]

[tex]\frac{X'}{X}=\lambda^{2} [/tex] and [tex]\frac{Y'}{Y}+1=\lambda^{2}[/tex]
 

Answers and Replies

  • #2
1,796
53
This is correct although normally for first order PDEs, we use the method of characteristics.
 

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