Engr 213 midterm 1a sol 2009
- 1. Concordia University February 13, 2009 Applied Differential Equations Section J Exam I (A) ANSWER KEY (1) (10 points) The ODE dy = 4(y 2 + 1) dx is separable. It can be written in the form dy = 4 dx, y2 +1 which, by integration leads to the solution in implicit form arctan y = 4x + C, where C is an arbitrary constant. (2) (10 points) The following equation dy + y = xy 4 dx is a Bernoulli equation with n = 4. We will use the substitution u = y −3 (with du = −3y −4 dy). Following this substitution, we obtain the linear ODE du − 3u = −3x. dx du An integrating factor is then µ(x) = e−3x , so e−3x − 3e−3x u = −3e−3x x, or dx (e−3x u) = −3xe−3x . Consequently, e−3x u = −3 xe−3x dx + C, or (using integration by parts) e−3x u = xe−3x − e−3x dx + C. 1 1 1 Finally, e−3x u = xe−3x + e−3x + C, so u = x + + Ce3x and y = (x + + Ce3x )−1/3 , 3 3 3 where C is an arbitrary constant. 1
- 2. 2 (3) Given x 3y 2 − x2 dx + + 2y dy = 0, 2y 4 y5 we consider f (x, y) such that ∂f x (x, y) = 4 ∂x 2y ∂f 3y 2 − x2 (x, y) = + 2y. ∂y y5 Integrating the first equation with respect to x, we obtain x2 f (x, y) = + c(y). 4y 4 To find c(y), we evaluate ∂f x2 3y 2 − x2 (x, y) = − 5 + c (y) = + 2y, ∂y y y5 3 3 (2y)3/2 hence c(y) satisfies the ODE: c (y) = + 2y. Therefore c(y) = − + + C, where y3 2y 2 3 C is an arbitrary constant. x2 3 (2y)3/2 Finally we have that f (x, y) = − 2+ + C, and we conclude that the general 4y 4 2y 3 solution of the given exact equation is (in implicit form) x2 3 (2y)3/2 − 2+ = c, where c is an arbitrary constant. 4y 4 2y 3 Now, y(0) = 2 implies c = 55/24, hence the solution of the IVP x2 3 (2y)3/2 55 − 2+ = . 4y 4 2y 3 24 (4) Denote by A(t) the number of pounds of salt in the tank at time t. Then we must solve the IVP dA 5A = 12 − , A(0) = 0. dt 100 − t 5 The ODE is linear with an integrating factor µ(t) = exp dt = exp(−5 ln(100 − 100 − t t)) = (100 − t)−5 . 12 So, (100−t)−5 A (t)+5A(100−t)−4 = 12(100−t)−5 , hence (100−t)−5 A(t) = (100−t)−4 +C 4 where C will be determined from A(0) = 0 to be equal to −3/1004 . 3 21 Finally, we have A(t) = 3(100 − t) − 4 (100 − t)5 and A(30) = 210 − 3 ≈ 209.97 lb. 100 10