Final Exam Solution - King Fahd University of Petroleum and Minerals
Final Exam Solution - King Fahd University of Petroleum and Minerals
Final Exam Solution - King Fahd University of Petroleum and Minerals
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Exercise 6 (10 points)a) Show that the functions f 1 = e x cos x <strong>and</strong> f 2 = e x sin x constitute a fundamentalset <strong>of</strong> solutions <strong>of</strong> the homogeneous differential equation 3y ′′ − 6y ′ + 6y = 0.b) Find the particular solution <strong>of</strong> the differential equation 3y ′′ − 6y ′ + 6y = e x sec x.<strong>Solution</strong>a) The wronskian <strong>of</strong> f 1 <strong>and</strong> f 2 is W =∣e x cosxe x cosx − e x sinxe x sinxe x sinx + e x cosx∣⇒ (1 point)W = e 2x sinxcosx + e 2x cos 2 x − e 2x sinxcosx + e 2x sin x = e 2x ≠ 0. ⇒ (1 point)Therefore f 1 <strong>and</strong> f 2 are two linearly independent solutions <strong>of</strong> the second order homogeneousdifferential equation. ⇒ (1 point)We deduce that f 1 <strong>and</strong> f 2 form a fundamental set <strong>of</strong> solutions. ⇒ (1 point)Total: (4 points)b) We use variation <strong>of</strong> parameters: y p = u 1 y 1 + u 2 y 2 . ⇒ (1 point)Let W 1 =∣0 e x sinx13 ex secx e x sinx + e x cosx∣ = − 1 3 e2x tanx. ⇒ (1 point)Then u ′ 1 = W 1W= − 1tanx. Therefore u 3 1 = 1 ln |cosx|. ⇒ (1 point)3Let W 2 =∣e x cosx 0e x cosx − e x sinx1 3 ex secx∣ = 1 3 e2x . ⇒ (1 point)Then u ′ 2 = W 2W= 1. Therefore u 3 2 = 1 x. ⇒ (1 point)3Thus y p = 1 3 ex (cosx ln |cosx| + xsinx). ⇒ (1 point)Total: (6 points)7
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