Differential equations (typical c4 test question)

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SUMMARY

The discussion centers on a differential equation representing the volume of liquid in a container, defined by the equation \(\frac{Dv}{dt} = 20 - kV\), where \(k\) is a positive constant. The equation reflects a constant inflow of 20 cm³/s and a proportional outflow based on the current volume. The solution to this differential equation is expressed as \(V = A + Be^{-kt}\), with \(A\) and \(B\) determined in terms of \(k\). The method of separation of variables is confirmed as the appropriate technique for solving this equation.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with the concept of separation of variables in calculus.
  • Knowledge of exponential functions and their properties.
  • Basic understanding of initial value problems in differential equations.
NEXT STEPS
  • Study the method of separation of variables in more depth.
  • Learn about first-order linear differential equations and their general solutions.
  • Explore the applications of differential equations in real-world scenarios.
  • Investigate the impact of varying the constant \(k\) on the solution behavior.
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Students studying calculus, particularly those focusing on differential equations, as well as educators looking for examples of first-order linear equations in practical applications.

thomas49th
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Homework Statement


Liquid is pouring into a container at a constant rate of 20 cm³/s and is leaking out at a rate proportional to the volume of liquid already in the container

a) Explain why, at t seconds, the volume, V cm³, of liquid in the conainer satisfies the differential equation:

\frac{Dv}{dt} = 20 - kV
where k is a positive constant

The container is initially empty

b) By solving the differential equation show that

V = A + Be^{-kt}

giving the values of A and B in terms of k

The Attempt at a Solution



a) Well the change in volume differs with time. 20 cm³/s is pouring in the the container minus a proportional rate of the volume in the container at the time.

b) NOT SURE

err do i have to do some flipping of differentials around?

Thanks :)
 
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b) yes... dv / (20-kV) = dt, then integrate and solve for V(t)
 
In other words, only separation of variables is needed to solve this DE.
 

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