Math 3200-001 Fall 2014 Practice exam 1 solutions 2/16/2014 Each problem is worth 0 to 4 points: 4=correct, 3=small error, 2=good progress, 1=some progress 0=nothing relevant. If the result is correct, detailed justification is not important though all steps need to be present so that your method is clear. The solutions below are at about the level of detail and justification I would like to see. 1. Find the general solution of the equation dy + 2xy = 0. dx Solution: By separation of variables, 1 y y = 2x y y dx = 2xdx, y 0 y dy = 2xdx ln y = x 2 + K y = ±e K e x2 = Ce x2 for any C, including C = 0, because y = 0 is also a solution: dy dx +2xy = 0 + x 0 = 0. 2. Solve dy = dt yet, y (0) = 2e Solution: Again by separation of variables, y y dt = e t dt, y 0 y dy = e t dt ln y = e t + K y = e et +K = Ce et which applies also for C = 0, because y = 0 is a solution of the differential equation. Applying the initial condition, Ce e0 = 2e Ce 1 = 2e C = 2 1
gives the solution y = 2e et. 3. Solve (1 + t) y + y = cos x, y (0) = 1, on the largest possible interval. Solution. Writing the differential equation in the form y + 1 1 + t y = cos t 1 + t, t 1 we see that it is a linear non-homogeneous first order equation, and because the coefficients are continuous on the interval ( 1, ) but not at the point 1, where the initial solution is given, the largest interval where the solution is guaranteed to exist and be unique is ( 1, ). (The solution might exist on a larger interval, but we do not know that yet.) Solving the homogeneous equation first, y + 1 y = 0, 1+t we have the y H = e 1 1+t dt = e ln 1+t +K = C 1 1 + t, which applies for C = 0 also because y = 0 is a solution of the homogeneous equation. We solve the nonhomogeneous equation on the interval (1, ), multiply it by the integrating factor (1 + t) y + y = cos t ((1 + t) y) = cos t µ = e ln(1+t) = 1 + t (1 + t) y = sin t + C y = sin t + C 1 + t and applying the initial condition y (0) = 1, we have 0 = y (0) = sin 0 + C 1 + 0 = C, thus y (t) = sin t, t ( 1, ). 1 + t Because lim t 1 + y (t) =, there is no way to extend y as a continuous function past the point 1, so ( 1, ) is indeed the largest interval where the solution exists. 2
4. Find the general solution of y = (4x + y) 2. Solution. Substitution u = 4x + y gives u = 4 + y u = u 2 + 4 which is separable, u u 2 + 4 dt = 1 du u 2 + 4 = dt 1 u 2 tan 1 2 = t + K u = 2 tan (2t + C). 5. Verify that the equation (1 + ye xy ) + (2y + xe xy ) y = 0 is exact and solve it. Solution. M = (1 + ye xy ), N = (2y + xe xy ) M/ y = xye xy N/ x = xye xy so the equation is exact. We are looking for a function H(x, y) such that H x = 1 + yexy, H y = 2y + xexy. Choose to integrate over x first, and get H (x, y) = 1 + ye xy dx = x + e xy + f (y). Subtituting in the second equation, gives 2y + xe xy = H y = y (x + exy + f (y)) = xe xy + f (y) so the solution is f (y) = 2y f (y) = y 2 + K H (x, y) = x + e xy + y 2 = C. 3
Test: H x = ( x + e xy + y 2) = 1 + ye xy x H y = ( x + e xy + y 2) = xe xy + 2y y 6. The time rate of change of a population P is proportional to P ( 1 P 1000), the population was 500 in year t = 0, then 600 in year t = 10. Find the formula for P (t). What are the population in year t = 20 and lim t P (t)? Solution. The solution of the logistic equation is Here we know that dp dt = rp P (t) = (1 PPL ), P (0) = P 0 P 0 P L P 0 (P 0 P L ) e rt. P (0) = 500, P (10) = 600, P L = 1000 and we need to determine r. It is easier and less error prone to proceed in general P (t) [ P 0 (P 0 P L ) e rt] = P 0 P L P 0 (P 0 P L ) e rt = P 0P L P (t) (P 0 P L ) e rt = P 0P L P (t) + P 0 = P 0 P (t) (P L P (t)) e rt = P 0 P (t) and then substitute specific numbers P L P (t) P L P 0 r = 1 t ln ( P0 P (t) e 10r = 500 1000 600 600 1000 500 = 2 3 r = 1 10 ln 2 ln 3 ln 2 =. 3 10 4 P L P (t) P L P 0 )
Sanity check: r > 0. Evaluate, P (20) = where e 20r = (e 10r ) 2 = ( 2 3) 2 so (500) (1000) 500 (500 1000) e 20r P (20) = (500) (1000) 500 (500 1000) ( ) 2 2 = 9000 13 3 Sanity check: 9000 13 700 > 500 and < 1000. Finally, lim P (t) = lim t t P 0 P L P 0 (P 0 P L ) e rt = 7. Solve the initial value problem ty + y = t 3 y 2, y (1) = 1. Solution. Dividing by t, the differential equation becomes y + t 1 y = t 2 y 2, t 0, P 0 P L P 0 (P 0 P L ) 0 = P 0P L P 0 = P L = 1000. so we solve in the interval (0, ) first. This equation is a Bernoulli differential equation y + p (t) y = q (t) y n, with n = 2. We know to use the substitution v = y 1 n = y 3, which gives v = y 3, v = 3y 2 y y = 1 3 y 2 v y 2 v + 3t 1 y = 3t 2 y 2 v + 3t 1 y 3 = 3t 2 v + 3t 1 v = 3t 2 which is a linear nonhomogeneous equation. Solve the homogeneous equation, v + 3t 1 v = 0 v = e 3t 1 dt = e 3 ln t +K = Ct 3 5
including C = 0 and absorbing the change of sign into C as well. The integrating factor is and for t > 0 we get µ (t) = e 3 t 1 dt = e 3 ln t = t 3, t 3 v + 3t 3 t 1 v = 3t 2 t 3 Going back to the original variable, (t 3 v) = 3t 5 t 3 v = 3t 5 dt = t6 2 + C v = t3 2 + C t 3. ( t y = v 1/3 3 = 2 + C ) 1/3. t 3 The initial condition y (1) = 1 gives ( 1 3 1 = y (1) = 2 + C ) 1/3 C = 1 2 ( 1 ( y (t) = t 3 + t 3)) 1/3. 2 Because lim t 0 + y (t) =, the solution cannot continue past t = 0. 8. Consider the equation for velocity of an object subject to gravity acceleration 9.81 ms 2 and falling with the drag linearly proportional to the velocity v according to the equation dv dt = g cv where the coefficient c = 5s 1. What will be the limit velocity lim t v (t)? 6
Solution. The homogeneous equation has the solution v = Ce ct, which gives the integrating factor µ (t) = e ct.then ct dv e dt + cvect = ge ct ( e ct v ) = ge ct e ct v = g c ect + C v = g c + Ce ct lim t v (t) = g c = 9.81ms 2 5s 1 = 9.81 5 ms 1. 7