4y" + (y') = 0
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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![### Differential Equation Problem
**Solve completely:**
\[4y'' + (y')^3 = 0\]
**Solution Steps:**
1. **Identify the Type of Equation:**
- This is a second-order differential equation where \( y'' \) represents the second derivative of \( y \) with respect to \( x \), and \( y' \) represents the first derivative of \( y \).
2. **Transformation to a Simpler Form:**
- To solve this, let's use \( v = y' \), where \( v \) is a function of \( y \). Then \( y'' = v \frac{dv}{dy} \).
- Substitute these into the given equation:
\[
4v \frac{dv}{dy} + v^3 = 0
\]
3. **Separation of Variables:**
- Factor out common terms and separate variables:
\[
v (4 \frac{dv}{dy} + v^2) = 0
\]
\[
v = 0 \quad \text{or} \quad 4 \frac{dv}{dy} = -v^2
\]
4. **Solving for \( v \):**
- If \( v = 0 \), \( y' = 0 \Rightarrow y = C_1 \).
- Otherwise:
\[
4 \frac{dv}{dy} = -v^2
\]
\[
\frac{4}{v^2} dv = -dy
\]
5. **Integrate Both Sides:**
- For the non-trivial solution:
\[
\int \frac{4}{v^2} dv = -\int dy
\]
\[
-\frac{4}{v} = y + C_2 \quad \Rightarrow \quad v = -\frac{4}{y + C_2}
\]
6. **Backward Transformation:**
- Substitute back \( v = y' \):
\[
y' = -\frac{4}{y + C_2}
\]
- Separate variables again for integration:
\[
(y + C_2) dy = -4 dx
\]
\[
\int (y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20e00ab3-e90c-493d-a4f6-cdfa310ac890%2F60c7680f-6ef3-4047-b5f1-d64ae410c18d%2Fkitx7hp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Differential Equation Problem
**Solve completely:**
\[4y'' + (y')^3 = 0\]
**Solution Steps:**
1. **Identify the Type of Equation:**
- This is a second-order differential equation where \( y'' \) represents the second derivative of \( y \) with respect to \( x \), and \( y' \) represents the first derivative of \( y \).
2. **Transformation to a Simpler Form:**
- To solve this, let's use \( v = y' \), where \( v \) is a function of \( y \). Then \( y'' = v \frac{dv}{dy} \).
- Substitute these into the given equation:
\[
4v \frac{dv}{dy} + v^3 = 0
\]
3. **Separation of Variables:**
- Factor out common terms and separate variables:
\[
v (4 \frac{dv}{dy} + v^2) = 0
\]
\[
v = 0 \quad \text{or} \quad 4 \frac{dv}{dy} = -v^2
\]
4. **Solving for \( v \):**
- If \( v = 0 \), \( y' = 0 \Rightarrow y = C_1 \).
- Otherwise:
\[
4 \frac{dv}{dy} = -v^2
\]
\[
\frac{4}{v^2} dv = -dy
\]
5. **Integrate Both Sides:**
- For the non-trivial solution:
\[
\int \frac{4}{v^2} dv = -\int dy
\]
\[
-\frac{4}{v} = y + C_2 \quad \Rightarrow \quad v = -\frac{4}{y + C_2}
\]
6. **Backward Transformation:**
- Substitute back \( v = y' \):
\[
y' = -\frac{4}{y + C_2}
\]
- Separate variables again for integration:
\[
(y + C_2) dy = -4 dx
\]
\[
\int (y
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