Final Exam Solution - King Fahd University of Petroleum and Minerals

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Exercise 3 (10 points)Find the solution of the following initial value problem( )( 1 −2−1X ′ =X, X(0) =3 −42).SolutionFirst we need to find the determinant of (A − λI).∣ 1 − λ −23 −4 − λ ∣ = (1 − λ)(−4 − λ) + 6 = λ2 + 3λ + 2. ⇒ (1 point)The characteristic equation is then λ 2 + 3λ + 2 = 0.The eigenvalues of A are λ 1 = −2 and λ 2 = −1. ⇒ (1 point)For λ 1 = −2:( ) ( )3 −2 | 0 3 −2 | 0(A − λI)K = 0 ⇒⇒⇒ (1 point)3 −2 | 0 0 0 | 0( 2)Let k 2 = r then k 1 = 2r and the eigenvector K 3 1 = r3. ⇒ (1 point)1For λ 2 = −1:(A − λI)K = 0 ⇒( 2 −2 | 03 −3 | 0) ( )2 −2 | 0⇒⇒ (1 point)0 0 | 0). ⇒ (1 point)( 1Let k 2 = s then k 1 = s and the eigenvector K 2 = s1Then the general solution is X(t) = c 1( 231)e −2t + c 2( 11)e −t ⇒ (1 point)To satisfy the initial condition, we get 2 3 c 1 + c 2 = −1 and c 1 + c + 2 = 2. ⇒ (1 point)Therefore c 1 = 9 and c 2 = −7.⇒ (1 point)( 6Finally, the solution of the IVP is X(t) =9) ( −7e −2t +−7)e −t ⇒ (1 point)4
Exercise 4 (10 points)Solve the following differential equation using an appropriate substitution:Solutiondydx = y2 + 2xyx 2 .Let y = ux, then dy = x du + u. ⇒ (2 points)dx dxSubstitution in the DE gives: x du + u = u2 x 2 +2ux 2dx x 2= u 2 + 2u ⇒ (1 point)Therefore du = u2 +udx xwhich is separable. ⇒ (1 point)We obtaindu= dx ⇒ ( 1 − 1 )du = dx . ⇒ (2 points)u(u+1) x u u+1 xThus ln |u| − ln |u + 1| = ln |x| + ln c ⇒ ln | u | = ln |cx|.⇒ (1 point)u+1Henceu= xc ⇒ 1 − xu+1 x+y= xc. ⇒ (2 points)Finally y = cx21−cx . ⇒ (1 point) 5
Page 1 and 2: Math 202- Elements of Differential
Page 3: Exercise 2 (12 points)Consider the
Page 7 and 8: Exercise 6 (10 points)a) Show that
Page 9 and 10: Exercise 8If we convert the Cauchy-
Page 11 and 12: Exercise 10The set {2x, x −1 , x
Page 13 and 14: Exercise 12Which one of the followi
Page 15: Exercise 14⎛Consider A = ⎝0 1

Final Exam Solution - King Fahd University of Petroleum and Minerals

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