Math 3200 Exam #1 Solutions
Briggs' Section - Spring 1998

1. 1. The ODE is second order and linear.
   2. Given the proposed solution y(x), we see that

      

      Substitution into the ODE results in

      

      as required.

   3. Applying the initial conditions to the general solution we find that

      

      Solving these two equations for A and B, we have A=1 and B=-1. So the general solution is .

2. The ODE is first order and nonlinear. Separating variables, we can write

   

   Integrating both sides leads to

   

   where C is an arbitrary constant that we redefined once. We can evaluate C by setting x=1 and y=4; this says that 2=1+C or C=1. Thus the solution to the initial value problem is .

3. 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. In this case, the function F must satisfy and . Note that the ODE is exact because . Integrating with respect to x, we have that , where g is an arbitrary functon of y. If we now differentiate this expression for F with respect to y, we have . Matching terms with , we see that which means (we can omit the arbitrary constant here). With g(y) determined, we have the solution given implicity by

   

   Check that which is the original ODE.

4. 1. Differentiating and substituting into the ODE, we have

      

      The only way this equation can be satisfied for all is if p( p-1)- p-8=0. This means that p satisfies the quadratic equation , or p=-2 or p=4. Therefore, and are solutions.

   2. We know that since this a second order ODE, the general solution must involve two arbitrary constants, and . Recall that we also (briefly) mentioned that a solution can be mutliplied by a constant or added to another solution and we still have a solution. So the most general way we can combine the two above solutions is ; this is the general solution.
5. While it's certainly acceptable, plotting points really is not necessary to solve this problem; a little ``qualitiative analysis'' works fine. The only direction field that has slopes near zero around the origin (x=0,y=0) is #1. The only ODE that has slopes near zero around the origin is d. So the first pair is certainly (1,d). Of the remmaining three direction fields, the only with negative slopes is #2. The only remaining ODE with negative slopes is b. So (2,b) is a pair. Direction field #4 has increasing slopes as which matches the behavior of ODE c. So (4,c) is a match. And (not by elimination!), direction field #3 has slopes aproaching zero as which matches the behavior of ODE a. So (3,a) is a match. In summary, we have (1,d), (2,b), (3,a), and (4,c).