Name                                        MATH 204 PRACTICE MIDTERM III   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 given differential equation by means of a power series about x = 0.  Find a recurrence relation and write the final solution in the form                 y'' + xy' + 2y  =  0,        y(0)  =  0,     y'(0)  =  1 Solution  We have         Plugging back into the original differential equation gives         Now adjust the sums so that they all contain the same powers of x.         Now pull out the n = 0 term and combine the sums         Next use the initial conditions to get         a0  =  0        a1  =  1         2a2 + 2a0  =  0   Hence          a2  =  0 The recursion relationship gives         (n + 2)(n + 1)an+2 + (n + 2)an  =  0 or                            -an             an+2  =                                                             n + 1 Hence all the even coefficients are 0.  The odds are                                       -1                      1                        -1                         1         a1  =  1        a3  =                a5  =                  a7  =                  a9  =                                                             2                    3. 4                    3. 4. 6                 3. 4. 6. 8 In general                              (-1)n                    (-1)n     a2n+1  =                                  =                                                     2. 4. 6. .... (2n)            2n n! The general solution is                                                          Problem 2  Determine the general solution of the differential equation that is valid in any interval not including the singular point.          x2y'' - xy' + y  =  0       Solution We let         y  =  xr Then          y'  =  rxr-1        y''  =  r(r - 1)xr-2 Substituting back into the original equation gives         r(r - 1)xr - rxr + xr  =  0         r2 - r - r + 1  =  r2 - 2r + 1  =  0         (r - 1)2  =  0         r  =  1 (repeated twice) Hence         y  =  (c1 + c2 ln|x|) |xr|                                                                                      Problem 3  Solve the following differential equation                         y(0)  =  y'(0)  =  0 Solution Since this is a piecewise defined differential equation, we use the method of Laplace Transforms.           L(y'') + 4L(y)  =  L(g)  =  L[sin t (1 - u2p(t))  =  L(sin t) - L[(sin t) u2p(t)]                                           1                     1         s2L(y) + 4L(y)  =              - e-2ps                                                       s2 + 1              s2 + 1 Solving for L(y), we get                                                 1         L(y)  =  (1 - e-2ps)                                                                       (s2 + 1)(s2 + 4) Now use partial fractions                     1                        As + B            Cs + D                                       =                      +                                    (s2 + 1)(s2 + 4)            s2 + 1             s2 + 4 or         (As + B)(s2 + 1) + (Cs + D)(s2 + 4)  =  1         As3 + Bs2 + As + B + Cs3 + Ds2 + 4Cs + 4D  =  0 Equating coefficients gives         A + C  =  0        B + D  =  0        A + 4C  =  0        B + 4D  =  0 This has solution          A  =  0        B  =  -1/3        C  =  0        D  =  1/3 Putting this together gives                        1                           1                   1          L(y)  =        (1 - e-2ps)                     -                                                3                       s2 + 4             s2 + 1                          1        1                   1                     e-2ps           e-2ps          L(y)  =        (                   -                  -                  +                )                        3       s2 + 4            s2 + 1            s2 + 4          s2 + 1  Now take the inverse Laplace Transform to get         y  =  1/3 ( sin(2t) - sin t - u2p sin(2t) + u2p sin t)                                                     Problem 4  Find the general solution of the given system of equations and describe the behavior of the solution as .             Solution Find the eigenvalues and eigenvectors of the matrix.                 =  (-2 - l)(-2 - l) - 1  =  l2 + 4l + 4 - 1         =  l2 + 4l + 3  =  (l + 3)(l + 1)  =  0 Thus the eigenvalues are          l  =  -3        and        l  =  -1 Now find the eigenvectors         v-3:  Plugging in -3 for l, we get         Which has (1,-1) in its null space.         v-1:  Plugging in -1 for l, we get         Which has (1,1) in its null space. The solution is          or         x1  =  c1e-3x + c2e-x         x2  =  -c1e-3x + c2e-x                                                    Problem 5  An electric circuit  is describes by the system of differential equations             A.     Suppose that R = 1 ohm, C = ½ farad, and L = 1 henry.  Find the general solution of the system in this case. Solution Plugging in the numbers we get the matrix        Now find the eigenvalues          Take the determinant to get          (-l)(-2 - l) + 2  =  l2 + 2l + 2  =  0 The roots of this quadratic are          -1 + i        and         -1 - i v-1+i:  Plugging in -1 + i for l, we get         The first row gives         (1 - i)x + y  =  0         y  =  (-1 + i)x So the corresponding eigenvector is          v-1+i  =  (1,-1 + i) We have         Hence the general solution is given by         That is         I(t)  =  c1e-t cost + c2e-t sin t         V(t)  =  c1e-t(-cos t - sin t) + c2e-t (cos t - sin t)                                        B.      Find I(t) and V(t) if I(0) = 2 amperes and V(0) = 1 volt. Plug in to get         0  =  c1                 1  =  -c1 + c2  The solution is          I(t)  =  e-t sin t         V(t)  =  e-t (cos t - sin t)                                 C.     For the circuit of part A. determine the limiting values of I(t) and V(t) as .  Do these limiting values depend on the initial conditions?   Solution Since the e-t term dominates all expressions, both the current and voltage will drop to zero as time approaches infinity independent of the initial conditions.                                   Problem 6  Please answer the following true or false.  If true, explain why.  If false, explain why or provide a counter-example.   A.    If  f(x)  is a function that is not continuous at x  =  2 , then the Laplace transform of  f(x)  is also not continuous at x =  2 .   Solution         False, consider u2(t) which is not continuous at 2.  The Laplace transform is e-2s / s which is continuous at 2.                               B.   Let      be a solution of the differential equation             x                     1                 y''  +                y'  + (sin x)y  =  0     x + 4              x - 2                  then  x  =  -1  is in the interval of convergence of y(x).   Solution False, the solution converges inside (0,2) which does not contain -1.