Find a solution to y" - 4y' - 5y = 2e²t using variation of parameters.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Please show steps I need help understanding how to use variation of parameters.
![**Problem 6: Applying Variation of Parameters**
Let's find a solution to the differential equation:
\[ y'' - 4y' - 5y = 2e^{2t} \]
using the method of variation of parameters.
### Step-by-Step Solution:
1. **Identify the Homogeneous Equation:**
The associated homogeneous equation is:
\[ y'' - 4y' - 5y = 0 \]
2. **Solve the Homogeneous Equation:**
The characteristic equation is:
\[ r^2 - 4r - 5 = 0 \]
Solving for \( r \):
\[
r = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2}
\]
Thus, \( r = 5 \) and \( r = -1 \).
The general solution to the homogeneous equation is:
\[
y_h(t) = C_1 e^{5t} + C_2 e^{-t}
\]
3. **Apply Variation of Parameters:**
The particular solution will take the form:
\[
y_p(t) = u_1(t) e^{5t} + u_2(t) e^{-t}
\]
Where \( u_1(t) \) and \( u_2(t) \) are functions to be determined.
4. **Determine \( u_1(t) \) and \( u_2(t) \):**
Find the Wronskian \( W \) of the homogeneous solutions \( y_1 = e^{5t} \) and \( y_2 = e^{-t} \):
\[
W = \begin{vmatrix}
e^{5t} & e^{-t} \\ 5e^{5t} & -e^{-t}
\end{vmatrix}
= e^{5t}(-e^{-t}) - e^{-t}(5e^{5t}) = -e^{4t} - 5e^{4t} = -6e^{4t}
\]
The formulas for \( u_1'(t) \) and \( u_2'(t) \) are:
\[
u_1'(t) = -\frac{y_2 g](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b491fec-b8bd-45cc-a06e-6176bd563445%2F3e7b5152-c4cc-42e4-b8b1-887370cfac6d%2Flyfll8s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 6: Applying Variation of Parameters**
Let's find a solution to the differential equation:
\[ y'' - 4y' - 5y = 2e^{2t} \]
using the method of variation of parameters.
### Step-by-Step Solution:
1. **Identify the Homogeneous Equation:**
The associated homogeneous equation is:
\[ y'' - 4y' - 5y = 0 \]
2. **Solve the Homogeneous Equation:**
The characteristic equation is:
\[ r^2 - 4r - 5 = 0 \]
Solving for \( r \):
\[
r = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm 6}{2}
\]
Thus, \( r = 5 \) and \( r = -1 \).
The general solution to the homogeneous equation is:
\[
y_h(t) = C_1 e^{5t} + C_2 e^{-t}
\]
3. **Apply Variation of Parameters:**
The particular solution will take the form:
\[
y_p(t) = u_1(t) e^{5t} + u_2(t) e^{-t}
\]
Where \( u_1(t) \) and \( u_2(t) \) are functions to be determined.
4. **Determine \( u_1(t) \) and \( u_2(t) \):**
Find the Wronskian \( W \) of the homogeneous solutions \( y_1 = e^{5t} \) and \( y_2 = e^{-t} \):
\[
W = \begin{vmatrix}
e^{5t} & e^{-t} \\ 5e^{5t} & -e^{-t}
\end{vmatrix}
= e^{5t}(-e^{-t}) - e^{-t}(5e^{5t}) = -e^{4t} - 5e^{4t} = -6e^{4t}
\]
The formulas for \( u_1'(t) \) and \( u_2'(t) \) are:
\[
u_1'(t) = -\frac{y_2 g
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