Engr 213 final sol 2009
- 1. Solutions to the Final exam of Winter 2009 Applied Ordinary Differential Equations ENGR 213 (1) (a) This is a separable equation which can be re-written in the form dy dx = . (2y + 3)2 (4x + 5)2 To integrate the left hand side, use the substitution u = 2y + 3, thus du = 2dy, while, for the right hand side, use w = 4x + 5, dw = 4dx. Therefore, dy 1 1 1 1 2 = 2 du = − + c = − + c, (2y + 3) 2 u 2u 2(2y + 3) dy 1 1 1 1 2 = 2 dw = − +c =− +c. (4x + 5) 4 w 4w 4(4x + 5) Using a single constant C to replace the difference c − c, we have the general solution of the equation in implicit form: 1 1 = + C. 2(2y + 3) 4(4x + 5) (b) This ODE is linear, hence we shall first put it in its standard form (we’ll solve it for x > 0): dy 1 1 + y = ex . dx x x x Thus P (x) = 1/x and Q(x) = e /x and we look for an integrating factor (hence in the following we choose the constant of integration zero) 1 dx µ(x) = e x = eln x = x. We multiply the equation with µ to obtain xy + y = ex or (xy) = ex . Integrating both sides, we obtain ex C xy = ex + C, C = constant, ⇒ y(x) = + , C = constant, x x the general solution in explicit form. We now make use of the initial condition y(1) = 2 to determine the constant: ex 2 − e y(1) = e + C = 2 ⇒ C = 2 − e ⇒ y(x) = + . x x (2) By denoting M (x, y) = y 2 cos x − 3x2 y − 2x, N (x, y) = 2y sin x − x3 + ln y, we check that ∂M ∂N (x, y) = (x, y) = 2y cos x − 3x2 ∂y ∂x hence the equation is exact. 1
- 2. 2 We therefore start to look for a function f (x, y) whose partial derivatives with respect to x, respectively y, are M and respectively N . We do so by integrating M (x, y) with respect to x: f (x, y) = y 2 sin x − x3 y − x2 + C(y). We then differentiate f with respect to y and, setting the result equal to N , obtain 2y sin x − x3 + C (y) = ln y. Thus C(y) is an antiderivative of ln y. We can find it by integration by parts (u = ln y, dv = dy): C(y) = ln y dy = y ln y − 1 dy = y ln y − y + c, c = constant. Hence f (x, y) = y 2 sin x − x3 y − x2 + y ln y − y + c and the solution to the ODE (in implicit form) is y 2 sin x − x3 y − x2 + y ln y − y + c = 0, c = constant. dy du (3) We’ll use the substitution u = x + y ⇒ y = u − x, then = − 1. Therefore, the dx dx substitution leads to the related ODE du du − 1 = tan2 u ⇔ = 1 + tan2 u. dx dx sin u sin2 u + cos2 u 1 As tan u = , we note that 1 + tan2 u = 2u = . Then cos u cos cos2 u du 1 = 2u ⇒ cos2 u du = dx dx cos 1 + cos(2u) u sin(2u) whose solution is du = x + c ⇒ + = x + c, thus the 2 2 4 general solution of the original ODE (in implicit form) is x + y sin(2(x + y)) + = x + c, c = constant. 2 4 (4) For the first 4 seconds, the velocity is constant (equal to 100 m/sec), then the velocity is a solution of the ODE dV = −0.002V 2 . dt This ODE has the solution 1 − = −0.002t + c, V (0) = 100 V or 1 1 V (t) = ⇒ = 20 ⇒ t = 20. 0.002t + 0.01 0.002t + 0.01
- 3. 3 We must not forget to add the 4 seconds when the velocity remained constant, hence the final answer is that the velocity reaches 20 m/sec after 24 sec. (5) (a) First, using r2 +6r +8 = 0 ⇔ r = −4 or r = −2, we find that the complementary part of the general solution is yc (x) = c1 e−4x + c2 e−2x . Having no duplication between yc and sin(3x), we set up yp as yp (x) = A sin(3x)+B cos(3x) ⇒ yp (x) = 3A cos(3x)−3B sin(3x) ⇒ yp (x) = −9A sin(3x)−9B cos(3x). Substituting these into the nonhomogenous ODE, we obtain −9A sin(3x) − 9B cos(3x) + 6(3A cos(3x) − 3B sin(3x)) + 8(A sin(3x) + B cos(3x)) = sin(3x), or −A − 18B = 1, 18A − B = 0 which implies A = −1/325 and B = −18/325. Thus the general solution of the ODE is 1 18 y(x) = c1 e−4x + c2 e−2x − sin(3x) − cos(3x), c1,2 = constants. 325 325 (b) We start again with the characteristic equation of the associated homogenous ODE, r2 +10r+25 = 0. This equation has 5 as a double root, hence yc (x) = c1 e5x +c2 xe5x . Yet, there is no duplication between yc and ex so we may take yp (x) = Aex = yp (x) = yp (x). In conclusion, 36Aex = ex , so A = 1/36 and the general solution of the ODE is 1 x y(x) = c1 e5x + c2 xe5x + e , c1,2 = constants. 36 (6) This is a non-homogeneous Cauchy-Euler equation. Consider first the homogeneous Cauchy-Euler DE: 2x2 y + 5xy + y = 0. Its characteristic equation is 2m(m − 1) + 5m + 1 = 0 or 2m2 + 3m + 1 = 0. Its roots are −1 and −1/2, hence (solving the ODE on (0, ∞)) 1 1 yc (x) = c1 + c2 √ , c1,2 = constants. x x By choosing, y1 (x) = x−1 and y2 (x) = x−1/2 , we can form the Wronskian W (x) and proceed to find a particular solution of the non-homogeneous equation given by using the variation of parameters. However, in order to apply the variation of parameters and identify correctly the function f needed for W1 and W2 below, we must use the standard form of the equation: y + P (x)y + Q(x) = f (x), thus divide the non-homogeneous Cauchy-Euler equation by 2x2 to obtain 5 1 1 1 y + y + 2y= − . 2x 2x 2 2x x−1 x−1/2 1 −5/2 0 x−1/2 1 1 W (x) = det = x , W1 (x) = det = − x−1/2 + x−3/2 −x −2 − 1 x−3/2 2 2 1 2 1 1 − 2x − 2 x−3/2 2 2 x−1 0 1 −1 1 −2 W2 (x) = det 1 1 = x − x , −x−2 2 − 2x 2 2
- 4. 4 W1 W2 u1 = = −x2 + x, u2 = = x3/2 − x1/2 . W W The simplest antiderivatives are x3 x2 2 2 u1 = − + , u2 = x5/2 − x3/2 . 3 2 5 3 Thus x3 x2 2 5/2 2 3/2 x2 x yp (x) = x−1 − + + x−1/2 x − x = − , 3 2 5 3 15 6 and 1 1 x2 x ygeneral (x) = c1 + c2 √ + − , c1,2 = constants. x x 15 6 (7) We’ll use the method of systematic elimination, hence, using D for the differentiation with respect to t, we’ll write the system as Dx = 2x + 3y − e2t , (D − 2)x − 3y = −e2t , or Dy = −x − 2y + e2t x + (D + 2)y = e2t . Through symbolic computation, we obtain (D − 2)(D + 2)x − 3(D + 2)y = −(D + 2)e2t (D2 − 4)x − 3(D + 2)y = −4e2t ⇒ 3x + 3(D + 2)y = 3e2t 3x + 3(D + 2)y = 3e2t . Above, we used that (D + 2)e2t = (e2t ) + 2e2t = 4e2t . Adding the two equations, we obtain (D2 − 1)x = −e2t ⇔ x − 1 = −e2t . We solve first this non-homogeneous ODE. It’s easy to see that xc (t) = c1 et + c2 e−t , c1,2 = constants. For the particular solution, we use the method of undetermined coefficients by setting xp (t) = Ae2t ⇒ xp (t) = 2Ae2t , xp (t) = 4Ae2t , thus 3A = −1 and 1 1 xp (t) = − e2t ⇒ x(t) = c1 et + c2 e−t − e2t , c1,2 = constants. 3 3 To find y(t) we go back to the first equation of the system and note that 1 c1 1 y = (x − 2x + e2t ) = − et − c2 e−t + e2t . 3 3 3 In conclusion, the general solution of the given system is x(t) = c1 et + c2 e−t − 1 e2t 3 c1,2 = constants. y(t) = 1 (x − 2x + e2t ) = − c31 et − c2 e−t + 1 e2t 3 3
- 5. 5 (8) Assume that the ODE has a solution of the form y(x) = an xn . Then y (x) = n≥0 n−1 n−2 nan x and y (x) = n(n − 1)an x . As y is a solution, we must have n≥1 n≥2 n(n − 1)an xn−2 − 3x nan xn−1 − an xn = 0. n≥2 n≥1 n≥0 Equivalently, n(n − 1)an xn−2 − 3nan xn − an xn = 0. n≥2 n≥1 n≥0 We’ll re-index the first sum as follows n(n − 1)an xn−2 = (n + 2)(n + 1)an+2 xn . n≥2 n≥0 Then (n + 2)(n + 1)an+2 xn − 3nan xn − a n xn = 0 n≥0 n≥1 n≥0 or 2a2 − a0 + ((n + 2)(n + 1)an+2 − 3nan − an ) xn = 0. n≥1 Thus a2 = a0 /2 and, generally, for n ≥ 1, (3n + 1)an an+2 = . (n + 2)(n + 1) Evaluating, we obtain 4 a3 = a1 3·2 7 7 a4 = a2 = a0 4·3 4·3·2 and so on. Thus 1 2 7 4 2 3 y(x) = a0 1+ x + x + . . . + a1 x+ x + ... 2 24 3 and it remains to find a0 and a1 such that y(0) = 1, y (0) = 0. Actually, y(0) = 1 ⇒ a0 = 1, while y (0) = 0 ⇒ a1 = 0, hence the solution is 1 2 7 4 y(x) = 1 + x + x + .... 2 24 (9) Note that this problem has been also solved in class so you may also refer to your lecture notes.
- 6. 6 (a) The ODE is q + 6q + 18q = 30 cos t, thus qc (t) = c1 e−3t cos(3t) + c2 e−3t sin(3t), c1,2 = constant. Setting up qp (t) = A cos t + B sin t, we obtain (−A + 6B + 18A) cos t + (−B − 6A + 18B) sin t = 30 cos t ⇒ B = 6A/17 A = 510/325 = 102/65 ⇒ B = 36/65. Thus the general solution is 102 36 q(t) = c1 e−3t cos(3t) + c2 e−3t sin(3t) + cos t + sin t, c1,2 = constant. 65 65 We see that 102 36 q (t) = c1 (−3e−3t cos(3t)−3e−3t sin(3t))+c2 (−3e−3t sin(3t)+3e−3t cos(3t))− sin t+ cos t, 65 65 so we can now use the initial conditions: q(0) = 100 = c1 + 102/65 and q (0) = 0 = −3c1 + 3c2 + 36/65 = 0. Thus 6398 6386 c1 = , c2 = , 65 65 so 6398 −3t 6386 −3t 102 36 q(t) = e cos(3t) + e sin(3t) + cos t + sin t. 65 65 65 65 (b) The transient terms are 6398 −3t 6386 −3t e cos(3t) and e sin(3t) 65 65 (as they go to zero as t approaches infinity) and the steady-state terms are the remaining (last) two. The circuit is underdamped.