MATH 204 PRACTICE MIDTERM I

Please work out each of the given problems on your own paper.  Credit will be based on the steps that you show towards the final answer.  Show your work.

Problem 1 

Solve the following differential equations.

A.     (25 points)                 x + y
                      y'  =                      
                                  x + 4y

Solution

This is a homogeneous differentiable equation.  Let

           v  =  y/x        y  =  xv        y'  =  v + xv'

                             x + xv                  1 + v
     v + xv'  =                      =                      
                        x + 4xv               1 + 4v

                         1 + v - v(1 + 4v)
      xv'  =                                             subtracting v and placing under a common denominator
                       1 + 4v                         

                         1 - 4v² 
         =                    
                   1 + 4v     

                      1 + 4v 
                                     dv    =  dx/x    Since v'  = dv/dx and then separate and factor.
           (1 - 2v)(1 + 2v)      

            Now use partial fractions

                      1 + 4v                             A                    B
                                         =                                                   
           (1 - 2v)(1 + 2v)                1 - 2v              1 + 2v

           A(1 + 2v)  + B(1 - 2v)  =  1 + 4v

            v  =  1/2:  2A  =  3        A  =  3/2

            v  =  -1/2:  2B  =  -1     B  =  -1/2

This yeilds

                3/2                -1/2
                        +                     dv    =  dx/x      
         1 - 2v            1 + 2v

Now integrate both sides to get

            -3/4 ln|1 - 2v| - 1/4 ln|1 + 2v|  =  ln|x| + C₁ 

             ln|(1 - 2v)³(1 + 2v)|  =  ln|x^-4| + C₂ 

             |(1 - 2v)³(1 + 2v)|  =  C x ^-4  

             |(1 - 2y/x)³(1 + 2y/x)|  =  C x ^-4  

             |(x - 2y)³(x + 2y)|  =  C  

B.     (25 points)    (e^y + e ^-x)dx + (e^y + 2ye^-x)dy  =  0

Solution

  We find an integrating factor

        M_y  =  e^y           N_x  =  -2y e ^-x 

             M_y - N_x             e^y + 2y e ^-x 
                          =                             =  1
             N                 e^y + 2y e ^-x      

We find the integrating factor by solving 

        dm/dx =  (1)m        m  =  e^x 

Now multiply the original differential equation by the integrating factor

         (e^x+y + 1)dx + (e^x+y + 2y)dy  =  0

This transformed differential equation is exact.  We can find the potential function.

        f_x  =  e^x+y + 1        

Integrate with respect to x to get

        f(x,y)  =  e^x+y + x + C(y)

Now take the partial derivative with respect to y to get

        f_y =  e^x+y + C'(y)  =  e^x+y + 2y

        C'(y)  =  2y        C(y)  =  y² 

Finally the solution is 

          e^x+y + x + y² = C

C.     (25 points)                             y
                      y'  =  A -                      
                                         B - Ct

  .
                                                        A(B - Ct)
            Solution:                    y  =                         + k(B - Ct)^1/C
                                                           1 - C

Solution

We can rewrite as

                          y 
       y'  +                 =  A
                  B - Ct

This is a first order linear differential equation.  We find the integrating factor

 Multiplying by the integrating factor gives

        (B - Ct)^-1/Cy' + (B - Ct)^{-1/C - 1}y  =  A(B - Ct)^-1/C 

        ((B - Ct)^-1/Cy)'  =  A(B - Ct)^-1/C

Now integrate both sides

  .
                           -A(B - Ct)^{-1/C + 1}                      AC(B - Ct)^{-1/C + 1}
(B - Ct)^-1/Cy  =                                    + k  =                                      + k                             
                                C(-1/C + 1)                                 1 - C

Now solve for y

                            A(B - Ct)
            y  =                        + k(B - Ct)^1/C
                          1 - C

Problem 2

Currently, Lower Angora Lake has 300 million gallons of unpolluted water in it.  Due to a gas leak in a motor boat in Upper Angora Lake, there is an inflow of 5000 gallons per hour that contains 2 grams per gallon of MTBE.  Since it is spring, the outflow of water from Lower Angora Lake is only 4000 gallons per hour.  Assume that the MTBE is completely mixed in the water as soon as it enters the lower lake.

A.     Set up a differential equation that describes the rate of change of MTBE in Lower Angora Lake.

Solution

This is 

        Rate  =  Rate In - Rate Out 

question.  First find the Rate In.

        Rate In  =  (5000gal/hr)(2grams/gal)  =  10,000 grams/hr

 Now find the Rate Out.  Notice that the number of gallons in the lake is given by

        Gallons =  300,000,000 + 1000t

We use this to calculate the Rate Out.

                                            x grams                           4000 gal
    Rate Out  =                                               ^.                          
                            300,000,000 + 1000t                 1  hr

Subtracting we get

             dx                                   4000 x grams          
                   =  10,000 -                                            
        dt                                 300,000,000 + 1000t       

B.     Use the result of Problem 1 Part C to determine the amount of MTBE in Lower Angora Lake one day after the leak began.

Solution

We have 

        A  =  10,000    B  =  300,000,000/4000 =  75,000    C  =  1000/4000  =  0.25

  .
                     10,000(75,000 - 0.25t)
          y  =                                               + k(75,000 - 0.25t)⁴
                      1 - 0.25

We find k by noticing that y(0)  =  0

            0  =  750,000,000/.75 + 75,000⁴k  =  10⁹ + 3.164 x 10¹⁹ k

            k  =  3.16 x 10^-11

Now plug in 1 for t to get 

       9999.93

Problem 3

The Tahoe area has enough resources to support up to 500 black bears, however if the bear population declines below 50, the bears will not be able to find mates and will eventually become extinct. 

A.  Set up an autonomous differential equation that models this situation.

Solution    

        y'  =  -y(1 - y/50)(1 - y/500)

B.  Sketch the direction field for this differential equation and sketch the solution that corresponds to the initial values:  y(0) = 45, y(0) = 100, and y(0) = 600. Assume the proportionality constant is 1.

   Solution

Problem 4

Please answer the following true or false.  If false, explain why or provide a counter-example.  If true, explain why.

A.      If a differential equation is autonomous then it is separable.

  Solution

True,  if

        dy/dx  =  f(y)

then 

        dy/f(y)  =  dx

separates the variable.

B.      The initial value problem y'  -  sin(t²) + e^t y  =  0,   y(1)  =  4  is guaranteed to have a unique solution defined for all values of t.

Solution

 True,  this is a first order linear differential equation with continuous p(t) and q(t).  Therefore there is a unique solution for all values of t.       

C.      The difference equation y_n+1  =  ky_n has solution y_n = y₀e^kn  

Solution

False, the solution is 

        y_n  =  y₀kⁿ

D.            d³y                       dy
     x            + ln(1 - x²)           =  x cos x
           dx³                       dx                          

   is a third order linear differential equation.

  Solution

True,  the highest derivative is y''', and the coefficients in front of y''' and y' are both functions of x.

Extra Credit:  Write down one thing that your instructor can do to make the class better and one thing that you want to remain the same in the class.

(Any constructive remark will be worth full credit.)