Engr 213 midterm 2a sol 2010
- 1. Concordia University March 23, 2010 Applied Ordinary Differential Equations ENGR 213 - Section F Prof. Alina Stancu Exam II (A) (1) (6 points) Solve the homogeneous ODE x2 y + 5xy + 4y = 0. Solution: The characteristic equation of this homogeneous Cauchy-Euler ODE is m(m − 1) + 5m + 4 = 0 or m2 + 4m + 4 = 0. It has a double root r = −2. Hence the general solution of the ODE on x > 0 (or x < 0) is y(x) = c1 x−2 + c2 x−2 ln x, c1,2 = constants. (2) (14 points) Solve the initial value problem y + 4y + 5y = e−2x , y(0) = 0, y (0) = 1. Solution: Consider first the associated homogeneous ODE: y + 4y + 5y = 0 with the characteristic equation r2 + 4r + 5 = 0 whose roots are r1,2 = −2 ± i. Hence yc (x) = c1 e−2x cos x + c2 e−2x sin x, c1,2 = constants. We now look for a particular solution yp to the non-homogenous ODE. We’ll use here the method of undetermined coefficients by setting yp (x) = Ae−2x . As yp (x) = −2Ae−2x and yp (x) = 4Ae−2x , we deduce that 4Ae−2x − 8Ae−2x + 5Ae−2x = e−2x ⇒ A = 1 and yp (x) = e−2x . Thus ygeneral (x) = c1 e−2x cos x + c2 e−2x sin x + e−2x , c1,2 = constants. We’ll now use the initial conditions to find c1,2 . As y(0) = 0, we have c1 + 1 = 0 ⇒ c1 = −1. Evaluating y (x) = c1 (−2e−2x cos x − e−2x sin x) + c2 (−2e−2x sin x + e−2x cos x) − 2e−2x , thus y (0) = −2c1 + c2 − 2 = c2 = 1. Therefore the solution of the IVP is y(x) = −e−2x cos x + e−2x sin x + e−2x . 1
- 2. 2 (3) (10 points) Use the variation of parameters to solve the differential equation y + y = cos2 x. Solution: The complementary part of the solution follows from r2 +1 = 0 ⇒ r = ±i and is yc (x) = c1 cos x + c2 sin x. Considering y1 (x) = cos x, y2 (x) = sin x, the Wronskian is W (x) = 1 = 0 for all real 0 sin x x’s. To find the complementary solution we calculate W1 (x) = det 2 = cos x cos x cos x 0 − sin x cos2 x and W2 (x) = det = cos3 x. − sin x cos2 x The method of variation of parameters gives yp (x) = y1 (x)u1 (x) + y2 (x)u2 (x), where u1 (x) = W1 (x)/W (x) and u2 (x) = W2 (x)/W (x). Integrating (by taking u = cos x du = − sin x dx) and taking the constant of integra- tion to be zero, we have u3 cos3 x u1 (x) = (− sin x cos2 x) dx = u2 du = = . 3 3 On the other hand, u3 sin3 x u2 (x) = cos3 x dx = cos x (1 − sin2 x) dx = (1 − u2 ) du = u − = sin x − , 3 3 where above we used the fundamental identity of trigonometry (sin2 x + cos2 x = 1) and the substitution u = sin x, du = cos x dx. Consequently, cos3 x sin3 x yp (x) = cos x · + sin x · sin x − 3 3 and cos4 x sin4 x y(x) = c1 cos x + c2 sin x + + sin2 x − , c1,2 = arbitrary constants. 3 3 (4) (10 points) A mass weighing 10 pounds stretches a spring 0.5 foot. Determine the equation of motion if the mass is initially released from a point 6 inches below the equilibrium position with a downward velocity of 4 ft/s. What is the instantaneous velocity at the first time when the mass passes through the equilibrium position? Solution: The equation of motion is mx + kx = 0, where m = 10/32 slug and k = 10/0.5 = 20 ft/lb. Thus x + 64x = 0 ⇒ x(t) = c1 cos 8t + c2 sin 8t, c1,2 = constants. To determine the constants, use the initial conditions. As x(0) = 1/2 ft, c1 = 1/2. Additionally, x (t) = −8c1 sin 8t + 8c2 cos 8t, thus x (0) = 4 ft/sec and 8c2 = 4.
- 3. 3 Consequently, x(t) = 1 cos 8t + 1 sin 8t and x (t) = −4 sin 8t + 4 cos 8t. To find the 2 2 time when the mass passes through the equilibrium position, set x(t) = 0. Note that this implies tan(8t) = −1 whose first positive solution is for 8t = 3π/4. So √ √ 2 2 √ x (3π/32) = −4 −4 = −4 2 ft/sec 2 2 is the answer needed.