Solve y" +10y' +24y = 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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![### Solving Second-Order Linear Homogeneous Differential Equations
In this module, we'll work through solving a second-order linear homogeneous differential equation. Consider the equation given below:
\[ y'' + 10y' + 24y = 0 \]
To solve this, we'll follow these steps:
1. **Characteristic Equation**: The first step is to form the characteristic (or auxiliary) equation by replacing \( y \) with \( e^{rt} \). This gives:
\[ r^2 + 10r + 24 = 0 \]
2. **Solve the Characteristic Equation**: Solve for \( r \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 10 \), and \( c = 24 \).
\[ r = \frac{-10 \pm \sqrt{(10)^2 - 4(1)(24)}}{2(1)} \]
\[ = \frac{-10 \pm \sqrt{100 - 96}}{2} \]
\[ = \frac{-10 \pm \sqrt{4}}{2} \]
\[ = \frac{-10 \pm 2}{2} \]
Thus, the solutions are:
\[ r = \frac{-10 + 2}{2} = -4 \]
\[ r = \frac{-10 - 2}{2} = -6 \]
3. **General Solution**: The general solution of the differential equation is a combination of the solutions derived from the characteristic equation, which are \( e^{-4t} \) and \( e^{-6t} \).
Hence, the general solution is:
\[ y(t) = C_1 e^{-4t} + C_2 e^{-6t} \]
Where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions.
### Summary
We've solved the differential equation \( y'' + 10y' + 24y = 0 \) by:
1. Forming the characteristic equation.
2. Solving the characteristic equation for \( r \).
3. Constructing the general solution using the roots of the characteristic equation.
Understanding this process can help in solving similar second-order linear homogeneous differential equations efficiently.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5dfcde0a-cc8e-4f7e-b5e2-9ec6ad36a3c9%2Fd9a767cf-9d3e-47b7-b58e-acbda5d49359%2Fk8q2h0fh_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Second-Order Linear Homogeneous Differential Equations
In this module, we'll work through solving a second-order linear homogeneous differential equation. Consider the equation given below:
\[ y'' + 10y' + 24y = 0 \]
To solve this, we'll follow these steps:
1. **Characteristic Equation**: The first step is to form the characteristic (or auxiliary) equation by replacing \( y \) with \( e^{rt} \). This gives:
\[ r^2 + 10r + 24 = 0 \]
2. **Solve the Characteristic Equation**: Solve for \( r \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 10 \), and \( c = 24 \).
\[ r = \frac{-10 \pm \sqrt{(10)^2 - 4(1)(24)}}{2(1)} \]
\[ = \frac{-10 \pm \sqrt{100 - 96}}{2} \]
\[ = \frac{-10 \pm \sqrt{4}}{2} \]
\[ = \frac{-10 \pm 2}{2} \]
Thus, the solutions are:
\[ r = \frac{-10 + 2}{2} = -4 \]
\[ r = \frac{-10 - 2}{2} = -6 \]
3. **General Solution**: The general solution of the differential equation is a combination of the solutions derived from the characteristic equation, which are \( e^{-4t} \) and \( e^{-6t} \).
Hence, the general solution is:
\[ y(t) = C_1 e^{-4t} + C_2 e^{-6t} \]
Where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions.
### Summary
We've solved the differential equation \( y'' + 10y' + 24y = 0 \) by:
1. Forming the characteristic equation.
2. Solving the characteristic equation for \( r \).
3. Constructing the general solution using the roots of the characteristic equation.
Understanding this process can help in solving similar second-order linear homogeneous differential equations efficiently.
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