Solve the given boundary-value problem. * Note 6-4ac =4-41:2= y" – 2y'+ 2y= 2x – 2, y(0)=0; y(1)= T sing (i) yc = Gem,X + Ca emax or C, Cos (A Sir Ithink this one (ii) yp=Ax+B y=ycʻyp->ylo)=0 y(r)=r -> C+ c
Solve the given boundary-value problem. * Note 6-4ac =4-41:2= y" – 2y'+ 2y= 2x – 2, y(0)=0; y(1)= T sing (i) yc = Gem,X + Ca emax or C, Cos (A Sir Ithink this one (ii) yp=Ax+B y=ycʻyp->ylo)=0 y(r)=r -> C+ c
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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![## Boundary Value Problem
### Problem Statement
Solve the given boundary-value problem:
\[ y'' - 2y' + y = 2x - 2, \quad y(0) = 0, \quad y(\pi) = \pi \]
### Solution Steps
#### Step 1: Solve the Homogeneous Equation
Using the homogeneous part of the equation:
\[ y'' - 2y' + y = 0 \]
**Characteristic Equation:**
\[ yc = y'' - 2y' + y = 0 \]
\[ yc = C_1 e^{m_1 x} + C_2 e^{m_2 x} \]
**Finding the Roots:**
\[ b^2 - 4ac = 4 - 4 \]
\[ 0 = 4 - 4 = 0 \]
**Therefore:**
\[ m = \alpha \pm i\beta \]
\[ yc = e^{\alpha x} \left[ C_1 \cos (\beta x) + C_2 \sin (\beta x) \right] \]
**Note:** Since \( b^2 - 4ac = 0 \), we have a repeated root. The general solution to the homogeneous equation will then be:
\[ yc = (C_1 + C_2 x) e^{\alpha x} \]
#### Step 2: Solve the Non-Homogeneous Equation (Particular Solution)
\[ yp = Ax + B \]
#### Step 3: General Solution
Combine the homogeneous (complementary) and particular solutions:
\[ y = yc + yp \]
#### Step 4: Apply Boundary Conditions
\[ y(0) = 0 \]
\[ y(\pi) = \pi \]
From this, solve for constants \( C_1 \) and \( C_2 \).
**Graphical Representation (If Any):**
If there were any graphs or diagrams, they would typically illustrate the solutions, characteristics equations, or functions \( y \) as they behave under given boundary conditions. However, none are included in the provided page.
**Conclusion:**
The general method to solve the given boundary-value problem involves finding solutions to both the homogeneous and particular parts of the equation, then combining them appropriately to satisfy the provided boundary conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c88e3a7-d6d9-4180-9e78-6a7df1a2f887%2Fce24035b-927d-4cd2-947a-5be7bf08011e%2Ff86p8sd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Boundary Value Problem
### Problem Statement
Solve the given boundary-value problem:
\[ y'' - 2y' + y = 2x - 2, \quad y(0) = 0, \quad y(\pi) = \pi \]
### Solution Steps
#### Step 1: Solve the Homogeneous Equation
Using the homogeneous part of the equation:
\[ y'' - 2y' + y = 0 \]
**Characteristic Equation:**
\[ yc = y'' - 2y' + y = 0 \]
\[ yc = C_1 e^{m_1 x} + C_2 e^{m_2 x} \]
**Finding the Roots:**
\[ b^2 - 4ac = 4 - 4 \]
\[ 0 = 4 - 4 = 0 \]
**Therefore:**
\[ m = \alpha \pm i\beta \]
\[ yc = e^{\alpha x} \left[ C_1 \cos (\beta x) + C_2 \sin (\beta x) \right] \]
**Note:** Since \( b^2 - 4ac = 0 \), we have a repeated root. The general solution to the homogeneous equation will then be:
\[ yc = (C_1 + C_2 x) e^{\alpha x} \]
#### Step 2: Solve the Non-Homogeneous Equation (Particular Solution)
\[ yp = Ax + B \]
#### Step 3: General Solution
Combine the homogeneous (complementary) and particular solutions:
\[ y = yc + yp \]
#### Step 4: Apply Boundary Conditions
\[ y(0) = 0 \]
\[ y(\pi) = \pi \]
From this, solve for constants \( C_1 \) and \( C_2 \).
**Graphical Representation (If Any):**
If there were any graphs or diagrams, they would typically illustrate the solutions, characteristics equations, or functions \( y \) as they behave under given boundary conditions. However, none are included in the provided page.
**Conclusion:**
The general method to solve the given boundary-value problem involves finding solutions to both the homogeneous and particular parts of the equation, then combining them appropriately to satisfy the provided boundary conditions.
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