Modeling with differential equations

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SUMMARY

The discussion focuses on solving the differential equation y' = -y^2. It establishes that y = 0 is a solution derived from setting -y^2 = 0. Additionally, it confirms that all members of the family y = 1/(x + C) are solutions, with the specific initial-value problem y' = -y^2 and y(0) = 0.5 leading to the solution y(x) = 1/(x + 1). The conversation emphasizes the importance of substituting functions into the equation to verify solutions.

PREREQUISITES
  • Understanding of differential equations, specifically first-order separable equations.
  • Familiarity with the concept of initial-value problems in calculus.
  • Knowledge of function substitution and differentiation techniques.
  • Basic skills in solving algebraic equations for constants.
NEXT STEPS
  • Study the method of separation of variables in differential equations.
  • Learn how to solve initial-value problems using specific examples.
  • Explore the implications of solution families in differential equations.
  • Investigate the behavior of solutions to nonlinear differential equations.
USEFUL FOR

Students studying calculus, particularly those focusing on differential equations, as well as educators seeking to enhance their teaching methods in this area.

bnwchbammer
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Homework Statement


a) What can you say about a solution of the equation y' = -y^2 just by looking at the differential equation?
b) Verify that all members of the family y = 1/(x + C) are solutions of the equation in part (a)
c) Can you think of a solution of the differential equation y' = -y^2 that is not a member of the family in part (b)?
d) Find a solution of the initial-value problem:
y' = -y^2 y(0) = 0.5


Homework Equations


Well for part b do I take the derivative of y? Other than that I don't believe there are specific equations.


The Attempt at a Solution


For part (a) I set -y^2 = 0 and got that a solution is y = 0. Not positive if that's what it's asking, however.
I'm not positive if I have to take the derivative in part (b), and I don't really know where to start for the rest.
 
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bnwchbammer said:

The Attempt at a Solution


For part (a) I set -y^2 = 0 and got that a solution is y = 0. Not positive if that's what it's asking, however.
What is not positive, you or y' ?
I'd say y' can't be posiitive so the solution are curves that are flat or slope downwards.
I'm not positive if I have to take the derivative in part (b)
The usual way to test if a specific function is a solution is to substitute it into the equation. That would involve taking the derivative when you substittue for y'
and I don't really know where to start for the rest.
c) You already thought of one solution Y = 0, although I don't think the way you derived it
made sense.

d) Use the form of the solution that they gave you, y(x) = 1/(x + C). Set x = 0 and solve for the value of C that makes y(0) = 0.5.
 

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