1. Solvedy - 5y = 2 using an Integrating Factor. an
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 First Order Linear Differential Equations using an Integrating Factor
**Problem:**
1. Solve \( \frac{dy}{dx} - 5y = 2 \) using an Integrating Factor.
**Solution:**
To solve the given first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), we use the integrating factor method.
1. **Identify \( P(x) \) and \( Q(x) \):**
\[
P(x) = -5, \quad Q(x) = 2
\]
2. **Find the integrating factor \( \mu(x) \):**
The integrating factor \( \mu(x) \) is given by:
\[
\mu(x) = e^{\int P(x) \, dx}
\]
Substituting \( P(x) \):
\[
\mu(x) = e^{\int -5 \, dx} = e^{-5x}
\]
3. **Multiply through by the integrating factor \( \mu(x) \):**
\[
e^{-5x} \frac{dy}{dx} - 5 e^{-5x} y = 2 e^{-5x}
\]
This can be rewritten using the product rule in reverse, recognizing that the left-hand side is the derivative of \( y \cdot e^{-5x} \):
\[
\frac{d}{dx} \left( y \cdot e^{-5x} \right) = 2 e^{-5x}
\]
4. **Integrate both sides:**
\[
\int \frac{d}{dx} \left( y \cdot e^{-5x} \right) dx = \int 2 e^{-5x} dx
\]
On the left side, the integral of the derivative gives:
\[
y \cdot e^{-5x} = \int 2 e^{-5x} dx
\]
Evaluate the right side integral:
\[
y \cdot e^{-5x} = 2 \left( \frac{e^{-5x}}{-5} \right) + C
\]
Simplify:
\[
y \cdot e^{-5x}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa70e8e95-a62f-4827-beb4-583f27efeab1%2Fb077e86c-c7a5-41bd-95f5-5c56552632d4%2F3arawml_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Solving First Order Linear Differential Equations using an Integrating Factor
**Problem:**
1. Solve \( \frac{dy}{dx} - 5y = 2 \) using an Integrating Factor.
**Solution:**
To solve the given first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), we use the integrating factor method.
1. **Identify \( P(x) \) and \( Q(x) \):**
\[
P(x) = -5, \quad Q(x) = 2
\]
2. **Find the integrating factor \( \mu(x) \):**
The integrating factor \( \mu(x) \) is given by:
\[
\mu(x) = e^{\int P(x) \, dx}
\]
Substituting \( P(x) \):
\[
\mu(x) = e^{\int -5 \, dx} = e^{-5x}
\]
3. **Multiply through by the integrating factor \( \mu(x) \):**
\[
e^{-5x} \frac{dy}{dx} - 5 e^{-5x} y = 2 e^{-5x}
\]
This can be rewritten using the product rule in reverse, recognizing that the left-hand side is the derivative of \( y \cdot e^{-5x} \):
\[
\frac{d}{dx} \left( y \cdot e^{-5x} \right) = 2 e^{-5x}
\]
4. **Integrate both sides:**
\[
\int \frac{d}{dx} \left( y \cdot e^{-5x} \right) dx = \int 2 e^{-5x} dx
\]
On the left side, the integral of the derivative gives:
\[
y \cdot e^{-5x} = \int 2 e^{-5x} dx
\]
Evaluate the right side integral:
\[
y \cdot e^{-5x} = 2 \left( \frac{e^{-5x}}{-5} \right) + C
\]
Simplify:
\[
y \cdot e^{-5x}
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