Solve the given initial value problem. y' + 8y' = 0; y(0) = - 10, y'(0) = 24 What is the auxiliary equation associated with the given differential equation? 2+8r=0 (Type an equation using r as the variable.) The solution is y(t) =
Solve the given initial value problem. y' + 8y' = 0; y(0) = - 10, y'(0) = 24 What is the auxiliary equation associated with the given differential equation? 2+8r=0 (Type an equation using r as the variable.) The solution is y(t) =
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 the Initial Value Problem**
The given initial value problem is:
\[ y'' + 8y' = 0; \quad y(0) = -10, \quad y'(0) = 24 \]
**Question:**
What is the auxiliary equation associated with the given differential equation?
**Step-by-Step Solution:**
1. **Formulate the Auxiliary Equation:**
Convert the given differential equation into its auxiliary equation form. For a differential equation of the form \( ay'' + by' + cy = 0 \), the corresponding auxiliary equation is:
\[ ar^2 + br + c = 0 \]
Here, the given differential equation is:
\[ y'' + 8y' = 0 \]
This translates to the auxiliary equation:
\[ r^2 + 8r = 0 \]
2. **Solving the Auxiliary Equation:**
To find the roots of the auxiliary equation, solve:
\[ r^2 + 8r = 0 \]
By factoring, we get:
\[ r(r + 8) = 0 \]
Thus, the roots are:
\[ r = 0, \quad r = -8 \]
3. **Formulate the General Solution:**
With roots \( r_1 \) and \( r_2 \) corresponding to a homogeneous linear differential equation, the general solution \( y(t) \) is given by:
\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]
Substituting the found roots:
\[ y(t) = C_1 e^{0 \cdot t} + C_2 e^{-8t} \]
Simplifying, we have:
\[ y(t) = C_1 + C_2 e^{-8t} \]
4. **Applying Initial Conditions:**
We use the initial conditions to solve for \( C_1 \) and \( C_2 \).
For \( y(0) = -10 \):
\[ -10 = C_1 + C_2 \cdot e^{0} \]
\[ -10 = C_1 + C_2 \]
For \( y'(0) = 24 \), take the derivative of \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe57d9078-3a26-4ab9-a5c6-f67f173c178d%2F56b40a57-d0ba-41f2-baba-7d1b3b5a86cd%2F8ab1nds_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving the Initial Value Problem**
The given initial value problem is:
\[ y'' + 8y' = 0; \quad y(0) = -10, \quad y'(0) = 24 \]
**Question:**
What is the auxiliary equation associated with the given differential equation?
**Step-by-Step Solution:**
1. **Formulate the Auxiliary Equation:**
Convert the given differential equation into its auxiliary equation form. For a differential equation of the form \( ay'' + by' + cy = 0 \), the corresponding auxiliary equation is:
\[ ar^2 + br + c = 0 \]
Here, the given differential equation is:
\[ y'' + 8y' = 0 \]
This translates to the auxiliary equation:
\[ r^2 + 8r = 0 \]
2. **Solving the Auxiliary Equation:**
To find the roots of the auxiliary equation, solve:
\[ r^2 + 8r = 0 \]
By factoring, we get:
\[ r(r + 8) = 0 \]
Thus, the roots are:
\[ r = 0, \quad r = -8 \]
3. **Formulate the General Solution:**
With roots \( r_1 \) and \( r_2 \) corresponding to a homogeneous linear differential equation, the general solution \( y(t) \) is given by:
\[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \]
Substituting the found roots:
\[ y(t) = C_1 e^{0 \cdot t} + C_2 e^{-8t} \]
Simplifying, we have:
\[ y(t) = C_1 + C_2 e^{-8t} \]
4. **Applying Initial Conditions:**
We use the initial conditions to solve for \( C_1 \) and \( C_2 \).
For \( y(0) = -10 \):
\[ -10 = C_1 + C_2 \cdot e^{0} \]
\[ -10 = C_1 + C_2 \]
For \( y'(0) = 24 \), take the derivative of \(
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