2. Find the general solution of d²y dy +5- dr.² dx 6y=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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![**Problem Statement:**
**2. Find the general solution of**
\[ \frac{d^2 y}{dx^2} + 5 \frac{dy}{dx} - 6y = 0 \]
**Detailed Solution:**
To find the general solution of the second-order linear homogeneous differential equation given above, follow these steps:
1. **Rewrite the Differential Equation**: The given differential equation is:
\[ \frac{d^2 y}{dx^2} + 5 \frac{dy}{dx} - 6y = 0 \]
2. **Find the Characteristic Equation**: Assume a solution of the form \( y = e^{rx} \). Substituting \( y = e^{rx} \), \( \frac{dy}{dx} = re^{rx} \), and \( \frac{d^2 y}{dx^2} = r^2 e^{rx} \) into the differential equation:
\[ r^2 e^{rx} + 5r e^{rx} - 6 e^{rx} = 0 \]
3. **Simplify and Solve for \(r\)**: Factor out \( e^{rx} \) which is never zero:
\[ e^{rx} (r^2 + 5r - 6) = 0 \]
So, we solve the characteristic equation:
\[ r^2 + 5r - 6 = 0 \]
4. **Solve the Quadratic Equation**: Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 5 \), and \( c = -6 \):
\[ r = \frac{-5 \pm \sqrt{25 + 24}}{2} \]
\[ r = \frac{-5 \pm \sqrt{49}}{2} \]
\[ r = \frac{-5 \pm 7}{2} \]
Thus we get two roots:
\[ r_1 = \frac{-5 + 7}{2} = 1 \]
\[ r_2 = \frac{-5 - 7}{2} = -6 \]
5. **Write the General Solution**: The general solution to the differential equation is a combination of the solutions corresponding to \( r_1 \) and \( r_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe02838e7-dc45-4607-88b3-6692c936110a%2F6d3ee6a5-d62f-449d-bc61-2ecf11b31265%2Fz531rg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
**2. Find the general solution of**
\[ \frac{d^2 y}{dx^2} + 5 \frac{dy}{dx} - 6y = 0 \]
**Detailed Solution:**
To find the general solution of the second-order linear homogeneous differential equation given above, follow these steps:
1. **Rewrite the Differential Equation**: The given differential equation is:
\[ \frac{d^2 y}{dx^2} + 5 \frac{dy}{dx} - 6y = 0 \]
2. **Find the Characteristic Equation**: Assume a solution of the form \( y = e^{rx} \). Substituting \( y = e^{rx} \), \( \frac{dy}{dx} = re^{rx} \), and \( \frac{d^2 y}{dx^2} = r^2 e^{rx} \) into the differential equation:
\[ r^2 e^{rx} + 5r e^{rx} - 6 e^{rx} = 0 \]
3. **Simplify and Solve for \(r\)**: Factor out \( e^{rx} \) which is never zero:
\[ e^{rx} (r^2 + 5r - 6) = 0 \]
So, we solve the characteristic equation:
\[ r^2 + 5r - 6 = 0 \]
4. **Solve the Quadratic Equation**: Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 5 \), and \( c = -6 \):
\[ r = \frac{-5 \pm \sqrt{25 + 24}}{2} \]
\[ r = \frac{-5 \pm \sqrt{49}}{2} \]
\[ r = \frac{-5 \pm 7}{2} \]
Thus we get two roots:
\[ r_1 = \frac{-5 + 7}{2} = 1 \]
\[ r_2 = \frac{-5 - 7}{2} = -6 \]
5. **Write the General Solution**: The general solution to the differential equation is a combination of the solutions corresponding to \( r_1 \) and \( r_
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