Solve the given initial-value problem. y(x) = y"" + 18y" + 81y' = 0, y(0) = 0, y'(0) = 1, y"(0) = -6
Solve the given initial-value problem. y(x) = y"" + 18y" + 81y' = 0, y(0) = 0, y'(0) = 1, y"(0) = -6
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 Initial-Value Problems: Example and Explanation ###
#### Example Problem ####
Solve the given initial-value problem.
\[
y''' + 18y'' + 81y' = 0, \quad y(0) = 0, \quad y'(0) = 1, \quad y''(0) = -6
\]
#### Solution ####
Here, we are provided with a third-order linear homogeneous differential equation:
\[ y''' + 18y'' + 81y' = 0 \]
with the associated initial conditions:
- \( y(0) = 0 \)
- \( y'(0) = 1 \)
- \( y''(0) = -6 \)
Our task is to find the function \( y(x) \) that satisfies both the differential equation and the initial conditions.
\[ y(x) = \, \boxed{ \quad \quad \quad \quad \quad \quad } \]
#### Explanation ####
1. **Characteristic Equation**:
To solve the given differential equation, we first find the characteristic equation associated with it. The characteristic equation is found by assuming a solution of the form \( y = e^{rx} \):
\[ r^3 + 18r^2 + 81r = 0 \]
2. **Finding Roots**:
Factorizing the characteristic equation:
\[ r(r^2 + 18r + 81) = 0 \]
\[ r(r + 9)^2 = 0 \]
Therefore, the roots are:
\[ r = 0, -9, -9 \]
3. **General Solution**:
Using the roots, the general solution to the differential equation is:
\[ y(x) = C_1 + C_2 e^{-9x} + C_3 xe^{-9x} \]
where \( C_1 \), \( C_2 \), and \( C_3 \) are constants that we determine using the initial conditions provided.
4. **Applying Initial Conditions**:
Initial condition: \( y(0) = 0 \)
\[ 0 = C_1 + C_2 + 0 \, \rightarrow C_1 + C_2 = 0 \]
Initial condition: \( y'(0) = 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa4d7cfc5-c108-459e-b939-f2d2e6c37b5b%2Ffc24a8f9-c43f-4bf6-b764-ffaba11baf2c%2Fuz11u09_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Initial-Value Problems: Example and Explanation ###
#### Example Problem ####
Solve the given initial-value problem.
\[
y''' + 18y'' + 81y' = 0, \quad y(0) = 0, \quad y'(0) = 1, \quad y''(0) = -6
\]
#### Solution ####
Here, we are provided with a third-order linear homogeneous differential equation:
\[ y''' + 18y'' + 81y' = 0 \]
with the associated initial conditions:
- \( y(0) = 0 \)
- \( y'(0) = 1 \)
- \( y''(0) = -6 \)
Our task is to find the function \( y(x) \) that satisfies both the differential equation and the initial conditions.
\[ y(x) = \, \boxed{ \quad \quad \quad \quad \quad \quad } \]
#### Explanation ####
1. **Characteristic Equation**:
To solve the given differential equation, we first find the characteristic equation associated with it. The characteristic equation is found by assuming a solution of the form \( y = e^{rx} \):
\[ r^3 + 18r^2 + 81r = 0 \]
2. **Finding Roots**:
Factorizing the characteristic equation:
\[ r(r^2 + 18r + 81) = 0 \]
\[ r(r + 9)^2 = 0 \]
Therefore, the roots are:
\[ r = 0, -9, -9 \]
3. **General Solution**:
Using the roots, the general solution to the differential equation is:
\[ y(x) = C_1 + C_2 e^{-9x} + C_3 xe^{-9x} \]
where \( C_1 \), \( C_2 \), and \( C_3 \) are constants that we determine using the initial conditions provided.
4. **Applying Initial Conditions**:
Initial condition: \( y(0) = 0 \)
\[ 0 = C_1 + C_2 + 0 \, \rightarrow C_1 + C_2 = 0 \]
Initial condition: \( y'(0) = 1
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