Math 3200 Exam #2
Spring 1998 - Briggs' Section

This is a 75-minute in-class exam. You are allowed to use a page of notes and a calculator. Please show and justify all of your work clearly.

  1. A patient in a hospital is receiving a drug intravenously at a rate of 25 milligrams per hour. The ODE that decribes the amount of drug in the patient's blood is y'(t)+0.04y(t) = 25. Assume that at t=0 there is no drug in the blood (y(0)=0).
    1. Find the solution to this initial value problem.
    2. What is tex2html_wrap_inline133 ? Describe what happens to the amount of drug in the blood as time increases.
    3. Suppose that the intravenous line is removed. What is the half-life of the drug?
  2. Consider the ODE

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    Find the general solution of this ODE and the solution that satisfies the condition y(1)=2.

  3. State whether and why each of the following sets of functions is a fundamental set for the ODE y''(x)-9y(x)=0.
    1. tex2html_wrap_inline139 .
    2. tex2html_wrap_inline141 .
    3. tex2html_wrap_inline143 .
    4. tex2html_wrap_inline145 .
  4. We studied the logistic equation as an example of a separable ODE. Solve the following initial value problem, treating the logistic equation as a Bernoulli equation.

    displaymath84

  5. The motion of a damped oscillator is governed by the initial value problem

    displaymath85

    Find the solution and make a rough sketch of the solution.

Math 3200 Exam #2 Solutions
Briggs' Section - Spring 1998

  1. We have a first order linear ODE to solve. The integrating factor is tex2html_wrap_inline147 . Multiplying the ODE by tex2html_wrap_inline149 gives us

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    Integrating both sides with respect to t leads to

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    where C is an arbitrary constant. Multiplying through by tex2html_wrap_inline155 gives us the general solution

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    The intitial condition y(0)=0 says that C=-625, so the solution of the initial value problem is

    displaymath89

    We see that as tex2html_wrap_inline161 , tex2html_wrap_inline163 which is the steady state amount of drug in the blood. Thus the amount of drug increases from an initial value of zero and approaches the asymptotic value of 625.

    If the input is stopped, this mens that the steady input of 25 on the right side of the ODE is replaced by zero: the ODE becomes y'(t)+0.04y(t)=0. This is the exponential decay equation with a rate constant of k=0.04. Thus the half-life of the drug is tex2html_wrap_inline169 hours. The half-life is the same regardless of how much drug is in the blood when the intravenous line is removed.

  2. The integrating factor for this first order linear ODE is tex2html_wrap_inline171 . Multiplying the ODE by tex2html_wrap_inline173 gives us

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    Integrating both sides with respect to x, we have

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    where C is an arbitrary constant. Multiplying through by tex2html_wrap_inline179 gives the general solution

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    The intitial condition y(1)=2 says that C=1, so the solution of the initial value problem is

    displaymath93

  3. Recall that a fundamental set is a set of two functions (since the ODE is second order) that are solutions of the ODE and that are linearly independent. The ODE was chosen so that with very little work you could see that the characteristic polynomial is tex2html_wrap_inline185 which has roots tex2html_wrap_inline187 . So two linearly independent solutions are tex2html_wrap_inline189 and tex2html_wrap_inline191 . Set (a) consists of constant multiples of these two linearly independent solutions, so it is a fundamental set. Set (b) has a function that is not a solution, so it is not a fundamental set. Set (c) consists of constant multiples of the two linearly independent solutions, so it is a fundamental set. Set (d) consists of constant multiples of the same solution, so it is not a linearly independent set and is not a fundamental set.
  4. If we rewite the ODE slightly, it takes the form of a Bernoulli equation

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    The necessary change of variables is tex2html_wrap_inline193 . It follows that tex2html_wrap_inline195 . If we divide the ODE through by tex2html_wrap_inline197 , it becomes

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    Substituting for v(t) gives the linear ODE

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    The integrating factor for this ODE is tex2html_wrap_inline201 . Multiplying through the ODE by this factor gives us

    displaymath97

    Integrating both sides with respect to t leads to

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    where C is an arbitrary constant. Multiplying through by tex2html_wrap_inline207 gives us the general solution

    displaymath99

    Transforming back to the original variable using y(t) = 1/v(t), we have the general solution for the original ODE:

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    Finally, the intitial condition y(0)=10 says that C=0.09, so the solution of the initial value problem is

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  5. The characteristic polynomial for this second order constant coefficient ODE is tex2html_wrap_inline215 . Using the quadratic formula, the roots are tex2html_wrap_inline217 . We can go immediaely to the general solution which is

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    For the initial conditions, we need the fact that

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    Then the initial conditions tell us that

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    The solution to the initial value problem is

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    The graph of the solution should show a damped oscillation (with period tex2html_wrap_inline219 ) within an envelope formed by the functions tex2html_wrap_inline221 .


Bill Briggs
Sun May 17 13:43:55 MDT 1998