Solve the initial-value problem d²y 4- - 4 - 3y = 0, y(0) = 1, y'(0) = 5. dx2 dy dx
Solve the initial-value problem d²y 4- - 4 - 3y = 0, y(0) = 1, y'(0) = 5. dx2 dy dx
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:
\[ 4 \frac{d^2y}{dx^2} - 4 \frac{dy}{dx} - 3y = 0, \]
with the initial conditions:
\[ y(0) = 1 \]
\[ y'(0) = 5 \]
### Steps to Solve the Initial-Value Problem
1. **Form the Auxiliary Equation:**
First, we convert the given differential equation into its corresponding auxiliary equation:
\[ 4r^2 - 4r - 3 = 0 \]
2. **Solve the Auxiliary Equation:**
Solve the quadratic equation \( 4r^2 - 4r - 3 = 0 \) to find the roots \(r\):
\[ r = \frac{4 \pm \sqrt{16 + 48}}{8} \]
\[ r = \frac{4 \pm \sqrt{64}}{8} \]
\[ r = \frac{4 \pm 8}{8} \]
So, \( r_1 = \frac{12}{8} = \frac{3}{2} \) and \( r_2 = \frac{-4}{8} = -\frac{1}{2} \).
3. **Form the General Solution:**
Based on the roots \( r_1 \) and \( r_2 \), the general solution of the differential equation is:
\[ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \]
\[ y(x) = C_1 e^{\frac{3}{2} x} + C_2 e^{-\frac{1}{2} x} \]
4. **Apply Initial Conditions:**
Use the initial conditions \( y(0) = 1 \) and \( y'(0) = 5 \) to find the constants \( C_1 \) and \( C_2 \):
- For \( y(0) = 1 \):
\[ 1 = C_1 e^{0} + C_2 e^{0} \]
\[ 1 = C_1 + C_2 \]
\[ C_1 + C_2 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f519245-0322-4f0e-8138-f81e9bcbc34d%2F79f80d1c-b68a-4df1-aea4-e1675be4d92b%2Ft3eybe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving the Initial-Value Problem
The given initial-value problem is:
\[ 4 \frac{d^2y}{dx^2} - 4 \frac{dy}{dx} - 3y = 0, \]
with the initial conditions:
\[ y(0) = 1 \]
\[ y'(0) = 5 \]
### Steps to Solve the Initial-Value Problem
1. **Form the Auxiliary Equation:**
First, we convert the given differential equation into its corresponding auxiliary equation:
\[ 4r^2 - 4r - 3 = 0 \]
2. **Solve the Auxiliary Equation:**
Solve the quadratic equation \( 4r^2 - 4r - 3 = 0 \) to find the roots \(r\):
\[ r = \frac{4 \pm \sqrt{16 + 48}}{8} \]
\[ r = \frac{4 \pm \sqrt{64}}{8} \]
\[ r = \frac{4 \pm 8}{8} \]
So, \( r_1 = \frac{12}{8} = \frac{3}{2} \) and \( r_2 = \frac{-4}{8} = -\frac{1}{2} \).
3. **Form the General Solution:**
Based on the roots \( r_1 \) and \( r_2 \), the general solution of the differential equation is:
\[ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \]
\[ y(x) = C_1 e^{\frac{3}{2} x} + C_2 e^{-\frac{1}{2} x} \]
4. **Apply Initial Conditions:**
Use the initial conditions \( y(0) = 1 \) and \( y'(0) = 5 \) to find the constants \( C_1 \) and \( C_2 \):
- For \( y(0) = 1 \):
\[ 1 = C_1 e^{0} + C_2 e^{0} \]
\[ 1 = C_1 + C_2 \]
\[ C_1 + C_2 =
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