Math 3200 Sample Exam #1
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. (10 points) Consider the ODE .
   1. Classify this ODE with respect to order (first, second, third...) and linearity (linear/nonlinear).
   2. For what values of m is the function a solution?
2. (10 points) Consider the ODE y''(x) + y'(x)-6y(x)=0.
   1. Classify this ODE with respect to order (first, second, third...) and linearity (linear/nonlinear).
   2. For what values of is the function a solution?
3. (10 points) Consider the first order nonlinear ODE

   

   1. Show that the ODE is exact.
   2. Find the general solution of the ODE, leaving it as an implicit function of x and y.
   3. Find a solution that satisfies the condition y(0)=0.
4. (10 points) A large cylindrical tank has a cross-sectional area of A square meters and a plugged hole in the bottom with an area of a square meters (see figure). At time t=0 the plug is removed and water gushes through the hole in the bottom of the tank. Let h(t) be the depth of water in the tank at time and assume that the initial depth of the water is h(0)=9 meters. The water depth is governed quite accurately by the ODE

   

   where g = 9.8 meters/sec is the acceleration of gravity. (You do not need to use the numerical value of g.)

   1. Assuming that A and a are given, solve this ODE for the depth function h(t). (Suggestions: simplify the notation by defining the constant , find the general solution, then use the initial condition.)
   2. Make a rough sketch of the solution using both intuition and the solution from part (a).
   3. How many seconds does it take to drain the tank?
5. (10 points) Use the integrating factor to solve the initial value problem

   

   Math 3200 - Solutions to Sample Exam #1

   1. The ODE is second order and linear. Differentiating and substituting into the ODE, we have

      

      The only way this equation can be satisfied for all is if m(m-1)+2m-6=0. This means that m satisfies the quadratic equation , or m=-3 or m=2. Therefore, and are solutions.

   2. The ODE is second order and linear. Differentiating and substituting into the ODE, we have

      

      Since , this equation is satisfied only if . This means that or . Therefore, and are solutions.

   3. Following the notation given in the book and in class, and M(x,y)=y(2x+y). Since , the equation is exact. Recall that with exact equations the goal is to find a function F(x,y) such that

      

      If this can be done, then a solution is given by F(x,y)=C, where C is an arbitrary constant. The function F must satisfy and . The first of these conditions means that . Integrating with respect to x, we have that , where g is an arbitrary functon of y. The second condition, implies that . Matching terms, we have that , or , where C is an arbitrary constant. With g(y) determined, we have the solution is given implicity by

      

      Note that the two arbitrary constants C can be combined. If we set x=0 and y=0, then we must take C=0. Therefore, the solution to the initial value problem (ODE plus initial condition) is .

   4. The ODE is separable. Integrating both sides with respect to t gives us

      

      where C is an arbitrary constant. Using the initial condition h(0)=9, we have . Therefore the solution is given by

      

      The tank is empty when h(t)=0 which happens when .

   5. Multiplying through the ODE by , we have

      

      Integrating both sides with respect to x gives us , where C is an arbitrary constant. The initial condition y(1)=8 implies that C=7. Therefore the solution is .