How Can You Solve the Differential Equation a=dv/dt=-Cv^2 for v(t) and x(t)?

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Homework Help Overview

The discussion revolves around solving the differential equation a = dv/dt = -Cv^2 for the functions v(t) and x(t), where C is a constant and the initial condition v(0) = v0 is provided. Participants are exploring methods of integration and the implications of the equation on velocity and distance.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to integrate both sides of the equation to express v(t) and x(t) but struggles with the integration of v^2(t). They also explore an alternative method but question its validity.
  • Some participants question the integration process and the role of a(t) in the integration of v^2(t), indicating confusion about the relationship between acceleration and velocity.
  • Another participant suggests the use of separation of variables as a potential approach to solving the differential equation.

Discussion Status

The discussion is ongoing, with participants seeking clarification on integration techniques and exploring different methods to approach the problem. There is a hint provided regarding separation of variables, which may guide further exploration.

Contextual Notes

Participants are working under the constraints of the problem as posed, including the initial condition and the nature of the differential equation. There is uncertainty regarding the integration of v^2(t) and the implications of the methods being considered.

dagr8est
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Given a(t) = -Cv^2(t) where C is a constant, express v(t) and x(t) as explicit functions of time t. Assume v(0) = v0

So I tried integrating both sides of the equation to get:
v(t)+C' = -C integral v^2(t) dt

but then I how am I supposed to integrate v^2(t) dt...

Also, I thought I could do it another way:

velocity
a(t) = -Cv^2(t)
v(t)/t = -Cv^2(t)
-1/(Ct) = v(t)

distance
-1/(Ct) = v(t)
-1/(Ct) = x(t)/t
-1/C = x(t)

Well the distance can't be constant so I guess that doesn't work either. I don't even know why this 2nd method doesn't work... It looks fine to me. Any help would be appreciated.
 
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You are correct in using integration to go from a(t) to an expression for v(t).

Are you just asking how to do the indefinite integral \int v^2 (t) dt ?
 
Yeah, I don't get how to integrate that. Wouldn't it be something like v^3(t)/(3a(t)) but where does the a(t) come from? It's not a constant so you just can't add it in right?
 
what??
notice that:
a=\frac{dv}{dt}=-Cv^2
how can you solve this differential equation?

hint: separation of variable!
 

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