2. Solve the following differential equation, showing all work. Verify the solution you obtain. y" - 2y' + y = 0; y(0) = 1, y'(0) = -2.
2. Solve the following differential equation, showing all work. Verify the solution you obtain. y" - 2y' + y = 0; y(0) = 1, y'(0) = -2.
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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![## Solving a Second-Order Differential Equation with Given Initial Conditions
2. **Problem Statement:**
Solve the following differential equation, showing all work. Verify the solution you obtain.
\[
y'' - 2y' + y = 0; \quad y(0) = 1, \quad y'(0) = -2.
\]
### Solution:
1. **Formulate the Characteristic Equation:**
For the given differential equation:
\[
y'' - 2y' + y = 0,
\]
we find the characteristic equation by assuming \( y = e^{rt} \). Substituting \( y = e^{rt} \) into the differential equation, we get:
\[
r^2 e^{rt} - 2r e^{rt} + e^{rt} = 0.
\]
Factoring out \( e^{rt} \) (which is never zero), we get:
\[
r^2 - 2r + 1 = 0.
\]
2. **Solve the Characteristic Equation:**
Solve this quadratic equation:
\[
r^2 - 2r + 1 = 0.
\]
This simplifies to:
\[
(r - 1)^2 = 0,
\]
giving a repeated root:
\[
r = 1.
\]
3. **Find the General Solution:**
Since \( r = 1 \) is a repeated root, the general solution to the differential equation is:
\[
y(t) = (C_1 + C_2 t) e^{rt} = (C_1 + C_2 t) e^{t}.
\]
4. **Apply the Initial Conditions:**
Use the initial conditions to solve for the constants \( C_1 \) and \( C_2 \):
\[
y(0) = 1 \Rightarrow (C_1 + C_2 \cdot 0) e^0 = C_1 = 1.
\]
Thus, \( C_1 = 1 \).
Next, differentiate the general solution to find \( y'(t) \):
\[
y'(t) = (C_1 + C_2 t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b491fec-b8bd-45cc-a06e-6176bd563445%2F0ee2bcf5-aa37-4b90-b942-dcee6f2790df%2Fjibu0wb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Solving a Second-Order Differential Equation with Given Initial Conditions
2. **Problem Statement:**
Solve the following differential equation, showing all work. Verify the solution you obtain.
\[
y'' - 2y' + y = 0; \quad y(0) = 1, \quad y'(0) = -2.
\]
### Solution:
1. **Formulate the Characteristic Equation:**
For the given differential equation:
\[
y'' - 2y' + y = 0,
\]
we find the characteristic equation by assuming \( y = e^{rt} \). Substituting \( y = e^{rt} \) into the differential equation, we get:
\[
r^2 e^{rt} - 2r e^{rt} + e^{rt} = 0.
\]
Factoring out \( e^{rt} \) (which is never zero), we get:
\[
r^2 - 2r + 1 = 0.
\]
2. **Solve the Characteristic Equation:**
Solve this quadratic equation:
\[
r^2 - 2r + 1 = 0.
\]
This simplifies to:
\[
(r - 1)^2 = 0,
\]
giving a repeated root:
\[
r = 1.
\]
3. **Find the General Solution:**
Since \( r = 1 \) is a repeated root, the general solution to the differential equation is:
\[
y(t) = (C_1 + C_2 t) e^{rt} = (C_1 + C_2 t) e^{t}.
\]
4. **Apply the Initial Conditions:**
Use the initial conditions to solve for the constants \( C_1 \) and \( C_2 \):
\[
y(0) = 1 \Rightarrow (C_1 + C_2 \cdot 0) e^0 = C_1 = 1.
\]
Thus, \( C_1 = 1 \).
Next, differentiate the general solution to find \( y'(t) \):
\[
y'(t) = (C_1 + C_2 t
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