Let ux,t=XxYy Given uxx+ux+uyy=Oux,y=XxYyux=X'xYyuxx=X"xYyuyy=XxY"y..uxx+ux+uyy=0⇒X"xYy+X'xYy+XxY"'y=0⇒X"x+X'XYy =-XXY"'y⇒X"x+X'xXx=-Y"yyy Let, X"X+X'XXX=-Y"yyy=X Case i: λ=0 Y"'y=0⇒Y'y=a⇒Yy=ay+b X"X+X'X=0 The characteristic function is D2+D=0⇒DD+1=0=D=0,-1 Therefore Xx=c+de-xux,t=a+byc+de-x Case ii: λ=µ2 Y"'y=-u2Yy⇒Yx=acosµy+bsinµyX"x+X'x-μ2xx=0 The characteristic function is D2+D-u2=0..D=-1+1+4u22 =-12+W w=1+4µ22 Xx=ce-12+wx+de-12-wxux,y=ce-12+wx+de-12- wxacosμy+bsinuy Case (iii): λ=-μ2 Y"'y=μ2Yy⇒Yx=aeµy+be-µyX"x+X'x+µ2Xx=0 D2+D-u2=0..D=-1±1-4µ22 For 1-4μ2>0
Let ux,t=XxYy Given uxx+ux+uyy=Oux,y=XxYyux=X'xYyuxx=X"xYyuyy=XxY"y..uxx+ux+uyy=0⇒X"xYy+X'xYy+XxY"'y=0⇒X"x+X'XYy =-XXY"'y⇒X"x+X'xXx=-Y"yyy Let, X"X+X'XXX=-Y"yyy=X Case i: λ=0 Y"'y=0⇒Y'y=a⇒Yy=ay+b X"X+X'X=0 The characteristic function is D2+D=0⇒DD+1=0=D=0,-1 Therefore Xx=c+de-xux,t=a+byc+de-x Case ii: λ=µ2 Y"'y=-u2Yy⇒Yx=acosµy+bsinµyX"x+X'x-μ2xx=0 The characteristic function is D2+D-u2=0..D=-1+1+4u22 =-12+W w=1+4µ22 Xx=ce-12+wx+de-12-wxux,y=ce-12+wx+de-12- wxacosμy+bsinuy Case (iii): λ=-μ2 Y"'y=μ2Yy⇒Yx=aeµy+be-µyX"x+X'x+µ2Xx=0 D2+D-u2=0..D=-1±1-4µ22 For 1-4μ2>0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Hello! Please, could you write this out the previous explanation on paper, because this is how I see it and it is very confusing, I don't understand, and I cannot follow along!
This is why I had specified in the beginning question about that!!
![The text appears to be a mathematical solution involving differential equations and the use of characteristic functions. Here is a transcription suitable for an educational website:
---
**Given:**
\[ u_{xx} + u_x + u_{yy} = 0 \]
Substitute variables:
\[ ux, t= XxYy \]
Rewriting, we have:
\[ y = XxYy; \quad u_{xx} + u_{yy} = 0 \Rightarrow X'' xY + X'xY' y + XxY'' y = 0 \]
This leads to:
\[ X'' + X' xX'' yY = \lambda \]
**Case i: \( \lambda = 0 \)**
\[ Y'' y = 0 \Rightarrow Y = a \Rightarrow Y = ay + b \]
\[ X'' x + X' x = 0 \]
The characteristic function becomes:
\[ D^2 + D = 0 \Rightarrow DD + 1 = 0 \Rightarrow D = 0, -1 \]
Therefore:
\[ Xx = c + de^{-x}, \quad t = a + byc + de^{-x} \]
**Case ii: \( \lambda = \mu^2 \)**
\[ Y'' y - \mu^2Y y = Yx = a \cos \mu y + b \sin \mu y \Rightarrow X'' x + X' x - \mu^2X x = 0 \]
The characteristic function is:
\[ D^2 + D - \mu^2 = 0 \Rightarrow D = -1 \pm 1 + 4 \mu^2 \]
Solving gives:
\[ -12 + w \quad \text{where} \quad w = 1 + 4 \mu^2 \]
Thus:
\[ Xx = ce^{-12 + wx} + de^{-12 - wx}UX, \quad y = ce^{-12 + wx} + de^{-12 - wx} a \cos \mu y + b \sin \mu y \]
**Case (iii): \( \lambda = -\mu^2 \)**
\[ Y'' y + 2Y y = Yx = ae^{\mu y} + be^{-\mu y} \Rightarrow X'' x + X' x + \mu^2 Xx =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76aaab3c-5a4a-4ff8-814d-d2e8a1f30217%2F70334b3f-27e4-47da-9082-e9ef123c73d7%2Ft2pgdkj_processed.png&w=3840&q=75)
Transcribed Image Text:The text appears to be a mathematical solution involving differential equations and the use of characteristic functions. Here is a transcription suitable for an educational website:
---
**Given:**
\[ u_{xx} + u_x + u_{yy} = 0 \]
Substitute variables:
\[ ux, t= XxYy \]
Rewriting, we have:
\[ y = XxYy; \quad u_{xx} + u_{yy} = 0 \Rightarrow X'' xY + X'xY' y + XxY'' y = 0 \]
This leads to:
\[ X'' + X' xX'' yY = \lambda \]
**Case i: \( \lambda = 0 \)**
\[ Y'' y = 0 \Rightarrow Y = a \Rightarrow Y = ay + b \]
\[ X'' x + X' x = 0 \]
The characteristic function becomes:
\[ D^2 + D = 0 \Rightarrow DD + 1 = 0 \Rightarrow D = 0, -1 \]
Therefore:
\[ Xx = c + de^{-x}, \quad t = a + byc + de^{-x} \]
**Case ii: \( \lambda = \mu^2 \)**
\[ Y'' y - \mu^2Y y = Yx = a \cos \mu y + b \sin \mu y \Rightarrow X'' x + X' x - \mu^2X x = 0 \]
The characteristic function is:
\[ D^2 + D - \mu^2 = 0 \Rightarrow D = -1 \pm 1 + 4 \mu^2 \]
Solving gives:
\[ -12 + w \quad \text{where} \quad w = 1 + 4 \mu^2 \]
Thus:
\[ Xx = ce^{-12 + wx} + de^{-12 - wx}UX, \quad y = ce^{-12 + wx} + de^{-12 - wx} a \cos \mu y + b \sin \mu y \]
**Case (iii): \( \lambda = -\mu^2 \)**
\[ Y'' y + 2Y y = Yx = ae^{\mu y} + be^{-\mu y} \Rightarrow X'' x + X' x + \mu^2 Xx =
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