Werify that (x)=- 1-ce* dy Substitute (x) for y and find the derivative to substitute for dx dy-y²-4y = dx A. 4 (1-ce)2 Simplify your answer.) = where c is an arbitrary constant, is a one-parameter family of solutions to dx (*) -A 1-ce* 1-ce* 4 4 The expression on the right can be further simplified by using a Choose the graph that shows the family of curves. of 4 Graph the solution curves corresponding to c = 0, +1, +2 using the same coord CEFE to subtract the fractions in the numerator, resulting in the same expression that was substitut OB.
Werify that (x)=- 1-ce* dy Substitute (x) for y and find the derivative to substitute for dx dy-y²-4y = dx A. 4 (1-ce)2 Simplify your answer.) = where c is an arbitrary constant, is a one-parameter family of solutions to dx (*) -A 1-ce* 1-ce* 4 4 The expression on the right can be further simplified by using a Choose the graph that shows the family of curves. of 4 Graph the solution curves corresponding to c = 0, +1, +2 using the same coord CEFE to subtract the fractions in the numerator, resulting in the same expression that was substitut OB.
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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![### Verification of Solution and Graphing Curves
#### Problem Statement
Verify that \( \phi(x) = \frac{4}{1 - ce^x} \), where \( c \) is an arbitrary constant, is a one-parameter family of solutions to
\[ \frac{dy}{dx} = \frac{y^2 - 4y}{4}. \]
Graph the solution curves corresponding to \( c = 0, \pm 1, \pm 2 \) using the same coordinate axes.
#### Solution Verification
1. Substitute \( \phi(x) \) for \( y \) and find the derivative to substitute for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{y^2 - 4y}{4} \]
\[ \frac{4}{(1 - ce^x)^2} = \frac{\left( \frac{4}{1 - ce^x} \right)^2 - 4 \left( \frac{4}{1 - ce^x} \right)}{4} \]
2. Simplify the expression:
\[ \frac{4}{(1 - ce^x)^2} = \frac{\frac{16}{(1 - ce^x)^2} - \frac{16}{1 - ce^x}}{4} \]
(Simplify your answer.) The expression on the right can further be simplified by using a common denominator to subtract the fractions in the numerator, resulting in the same expression that was substituted for \( \frac{dy}{dx} \) on the left.
#### Choose the Correct Graph
Choose the graph that shows the family of curves for the specific values of \( c \).
- **Option A**: Graph with curves labeled \( c = 2, 1, 0, -1, -2 \).
- **Option B**: Graph with curves labeled \( c = 2, 1, 0, -1, -2 \).
- **Option C**: Graph with curves labeled \( c = 2, 1, 0, -1, -2 \).
- **Option D**: Graph with curves labeled \( c = 1, 0, -1, 2 \).
Based on the visual inspection of the provided graphs:
- **Option A** depicts a set of hyperbolic-like curves.
- **](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc45acf51-cefc-42df-83e4-535a253da58c%2Fbc7566d0-5e0a-4620-bb09-09aaefbfcd9c%2F61bkpl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Verification of Solution and Graphing Curves
#### Problem Statement
Verify that \( \phi(x) = \frac{4}{1 - ce^x} \), where \( c \) is an arbitrary constant, is a one-parameter family of solutions to
\[ \frac{dy}{dx} = \frac{y^2 - 4y}{4}. \]
Graph the solution curves corresponding to \( c = 0, \pm 1, \pm 2 \) using the same coordinate axes.
#### Solution Verification
1. Substitute \( \phi(x) \) for \( y \) and find the derivative to substitute for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{y^2 - 4y}{4} \]
\[ \frac{4}{(1 - ce^x)^2} = \frac{\left( \frac{4}{1 - ce^x} \right)^2 - 4 \left( \frac{4}{1 - ce^x} \right)}{4} \]
2. Simplify the expression:
\[ \frac{4}{(1 - ce^x)^2} = \frac{\frac{16}{(1 - ce^x)^2} - \frac{16}{1 - ce^x}}{4} \]
(Simplify your answer.) The expression on the right can further be simplified by using a common denominator to subtract the fractions in the numerator, resulting in the same expression that was substituted for \( \frac{dy}{dx} \) on the left.
#### Choose the Correct Graph
Choose the graph that shows the family of curves for the specific values of \( c \).
- **Option A**: Graph with curves labeled \( c = 2, 1, 0, -1, -2 \).
- **Option B**: Graph with curves labeled \( c = 2, 1, 0, -1, -2 \).
- **Option C**: Graph with curves labeled \( c = 2, 1, 0, -1, -2 \).
- **Option D**: Graph with curves labeled \( c = 1, 0, -1, 2 \).
Based on the visual inspection of the provided graphs:
- **Option A** depicts a set of hyperbolic-like curves.
- **
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