Verify that y = ODE y" + 3y' + 2y = 4x². -4e-x-2e-2x + 2x² - 6x +7 is a solution of the SHOW ALL WORK!
Verify that y = ODE y" + 3y' + 2y = 4x². -4e-x-2e-2x + 2x² - 6x +7 is a solution of the SHOW ALL WORK!
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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![### Verifying the Solution of a Second-Order Ordinary Differential Equation (ODE)
Given the ODE:
\[ y'' + 3y' + 2y = 4x^2 \]
We need to verify that:
\[ y = -4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7 \]
is a solution.
#### Step-by-Step Solution:
1. **First Derivative (\( y' \))**:
Given \( y \):
\[ y = -4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7 \]
Compute \( y' \):
\[
y' = \frac{d}{dx} \left( -4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7 \right)
\]
\[
y' = 4e^{-x} + 4e^{-2x} + 4x - 6
\]
2. **Second Derivative (\( y'' \))**:
Given \( y' \):
\[ y' = 4e^{-x} + 4e^{-2x} + 4x - 6 \]
Compute \( y'' \):
\[
y'' = \frac{d}{dx} \left( 4e^{-x} + 4e^{-2x} + 4x - 6 \right)
\]
\[
y'' = -4e^{-x} - 8e^{-2x} + 4
\]
3. **Substitute \( y \), \( y' \), and \( y'' \) into the ODE**:
\[
y'' + 3y' + 2y = 4x^2
\]
Substitute the computed values:
\[
(-4e^{-x} - 8e^{-2x} + 4) + 3(4e^{-x} + 4e^{-2x} + 4x - 6) + 2(-4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7)
\]
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5dfcde0a-cc8e-4f7e-b5e2-9ec6ad36a3c9%2Ff11bc9ad-18d5-450f-8d02-7c642ebdf81c%2F0e6uhqzr_processed.png&w=3840&q=75)
Transcribed Image Text:### Verifying the Solution of a Second-Order Ordinary Differential Equation (ODE)
Given the ODE:
\[ y'' + 3y' + 2y = 4x^2 \]
We need to verify that:
\[ y = -4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7 \]
is a solution.
#### Step-by-Step Solution:
1. **First Derivative (\( y' \))**:
Given \( y \):
\[ y = -4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7 \]
Compute \( y' \):
\[
y' = \frac{d}{dx} \left( -4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7 \right)
\]
\[
y' = 4e^{-x} + 4e^{-2x} + 4x - 6
\]
2. **Second Derivative (\( y'' \))**:
Given \( y' \):
\[ y' = 4e^{-x} + 4e^{-2x} + 4x - 6 \]
Compute \( y'' \):
\[
y'' = \frac{d}{dx} \left( 4e^{-x} + 4e^{-2x} + 4x - 6 \right)
\]
\[
y'' = -4e^{-x} - 8e^{-2x} + 4
\]
3. **Substitute \( y \), \( y' \), and \( y'' \) into the ODE**:
\[
y'' + 3y' + 2y = 4x^2
\]
Substitute the computed values:
\[
(-4e^{-x} - 8e^{-2x} + 4) + 3(4e^{-x} + 4e^{-2x} + 4x - 6) + 2(-4e^{-x} - 2e^{-2x} + 2x^2 - 6x + 7)
\]
4
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