Verify for a > 0, that y = is a solution of the nonlinear differential equation yyll = x.
Verify for a > 0, that y = is a solution of the nonlinear differential equation yyll = x.
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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![**Problem:**
Verify for \( x > 0 \), that \( y = \frac{2}{\sqrt{3}} x^{3/2} \) is a solution of the nonlinear differential equation \( y y'' = x \).
**Solution:**
Steps to verify the solution:
1. **Find the Second Derivative**:
- First, find \( y' \) (the first derivative of \( y \)).
- Then, find \( y'' \) (the second derivative of \( y \)).
2. **Substitute \( y \) and \( y'' \)** into the given differential equation \( y y'' = x \).
3. **Validate** the equality to confirm that the solution satisfies the equation for \( x > 0 \).
Let's derive \( y' \) and \( y'' \) from \( y = \frac{2}{\sqrt{3}} x^{3/2} \):
- First derivative:
\[
y = \frac{2}{\sqrt{3}} x^{3/2}
\]
\[
y' = \frac{d}{dx}\left(\frac{2}{\sqrt{3}} x^{3/2}\right) = \frac{2}{\sqrt{3}} \cdot \frac{3}{2} x^{(3/2)-1} = \frac{3}{\sqrt{3}} x^{1/2} = \sqrt{3} x^{1/2}
\]
- Second derivative:
\[
y' = \sqrt{3} x^{1/2}
\]
\[
y'' = \frac{d}{dx} (\sqrt{3} x^{1/2}) = \sqrt{3} \cdot \frac{1}{2} x^{(1/2)-1} = \frac{\sqrt{3}}{2} x^{-1/2}
\]
Now, substitute \( y \) and \( y'' \) back into the original differential equation \( y y'' = x \):
- Given:
\[
y y'' = \frac{2}{\sqrt{3}} x^{3/2} \cdot \frac{\sqrt{3}}{2} x^{-1/2}
\]
\[
y y'' = \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc262d38-8642-4292-8f86-e10c24c29ddb%2F4786f717-2ec5-4a50-bf60-45d10d750a52%2Flise8am_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
Verify for \( x > 0 \), that \( y = \frac{2}{\sqrt{3}} x^{3/2} \) is a solution of the nonlinear differential equation \( y y'' = x \).
**Solution:**
Steps to verify the solution:
1. **Find the Second Derivative**:
- First, find \( y' \) (the first derivative of \( y \)).
- Then, find \( y'' \) (the second derivative of \( y \)).
2. **Substitute \( y \) and \( y'' \)** into the given differential equation \( y y'' = x \).
3. **Validate** the equality to confirm that the solution satisfies the equation for \( x > 0 \).
Let's derive \( y' \) and \( y'' \) from \( y = \frac{2}{\sqrt{3}} x^{3/2} \):
- First derivative:
\[
y = \frac{2}{\sqrt{3}} x^{3/2}
\]
\[
y' = \frac{d}{dx}\left(\frac{2}{\sqrt{3}} x^{3/2}\right) = \frac{2}{\sqrt{3}} \cdot \frac{3}{2} x^{(3/2)-1} = \frac{3}{\sqrt{3}} x^{1/2} = \sqrt{3} x^{1/2}
\]
- Second derivative:
\[
y' = \sqrt{3} x^{1/2}
\]
\[
y'' = \frac{d}{dx} (\sqrt{3} x^{1/2}) = \sqrt{3} \cdot \frac{1}{2} x^{(1/2)-1} = \frac{\sqrt{3}}{2} x^{-1/2}
\]
Now, substitute \( y \) and \( y'' \) back into the original differential equation \( y y'' = x \):
- Given:
\[
y y'' = \frac{2}{\sqrt{3}} x^{3/2} \cdot \frac{\sqrt{3}}{2} x^{-1/2}
\]
\[
y y'' = \
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