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midterm152sol

midterm152sol - Part I Answer each of the following 16...

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Part I: Answer each of the following 16 multiple choice questions. Each correct answer is worth 2 marks. Question 1 2 3 4 5 6 7 8 9 10 Answer (d) (e) (d) (b) (c) (d) (e) (c) (c) (c) Question 11 12 13 14 15 16 Total Answer (d) (b) (a) (e) (c) (e) 1. The function y = e - t is a solution of the diﬀerential equation (a) 2 y 000 - 2 y 00 - 9 y 0 + 9 y = 0 (b) y 00 + y 0 = 2 e - t (c) 4 t 2 y 00 + 8 ty 0 + y = 0 (d) ty 00 + ( t + 1) y 0 + y = 0 (e) y 00 + y = 0 Solution : (a) 2 y 000 - 2 y 00 - 9 y 0 + 9 y = 2( - e - t ) - 2( e - t ) - 9( - e - t ) + 9( e - t ) = 14 e - t 6 = 0 . (b) y 00 + y 0 - 2 e - t = ( e - t ) + ( - e - t ) - 2 e - t = - 2 e - t 6 = 0 . (c) 4 t 2 y 00 + 8 ty 0 + y = 4 t 2 ( e - t ) + 8 t ( - e - t ) + ( e - t ) = (4 t 2 - 8 t + 1) e - t 6 = 0 . (d) ty 00 + ( t + 1) y 0 + y = t ( e - t ) + ( t + 1)( - e - t ) + ( e - t ) = 0 . (e) y 00 + y = ( e - t ) + ( e - t ) = 2 e - t 6 = 0 . 2. Find a fundamental set of solutions for 2 y 00 + 5 y 0 - 3 y = 0 . (a) e 1 2 t , e 3 t (b) e - 3 t , e 2 t (c) e - 3 t , e - 1 2 t (d) e 1 2 t , 1 (e) e 1 2 t - e - 3 t , e 1 2 t + e - 3 t Solution : The characteristic equation is 2 λ 2 + 5 λ - 3 = (2 λ - 1)( λ + 3) = 0 which gives two distinct real roots λ 1 = 1 2 and λ 2 = - 3. The diﬀerential equation has the solutions y 1 = e 1 2 t , y 2 = e - 3 t . Since the given equation is linear homogeneous, the principle of superposition implies that c 1 y 1 + c 2 y 2 is again a solution for any choice of c 1 , c 2 . In particular, y 1 - y 2 and y 1 + y 2 are solutions. Hence, the pair of functions e 1 2 t - e - 3 t , e 1 2 t + e - 3 t can form a fundamental set of solutions because both are solutions and they are actually linearly independent.

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1 3. Find the integrating factor that can be used to solve the following ﬁrst-order equation dr + r tan θ = cos 2 θ, 0 < θ < π 2 . (a) sin θ (b) cos θ (c) tan θ (d) sec θ (e) csc θ Solution : The integrating factor is given by μ ( θ ) = exp •Z tan θ dθ = exp •Z sin θ cos θ = exp - Z 1 cos θ d (cos θ ) = exp[ - ln(cos θ )] = exp ln( 1 cos θ ) = 1 cos θ = sec θ. 4. The following equation is not exact but becomes exact when multiplied by the integrating factor. ( 3 xy + y 2 ) dx + ( x 2 + xy ) dy = 0 . (a) 2 (b) x (c) y (d) e x (e) x 2 Solution : By multiplying the factor x to the given equation, we have x ( 3 xy + y 2 ) dx + x ( x 2 + xy ) dy = 0 , or in the form of M ( x,y ) dx + N ( x,y ) dy = 0, where M ( x,y ) = 3 x 2 y + xy 2 , N ( x,y ) = x 3 + x 2 y. Then the new equation becomes exact because ∂M ∂y = ∂y £ 3 x 2 y + xy 2 / = 3 x 2 + 2 xy, ∂N ∂x = ∂x £ x 3 + x 2 y / = 3 x 2 + 2 xy, that is, the condition ∂M ∂y = ∂N ∂x is satisﬁed.
2 5. The following system of ﬁrst-order equations ( x 0 1 = - 3 x 1 + 7 x 2 , x 1 (0) = 2 , x 0 2 = 2 x 1 - 4 x 2 , x 2 (0) = 1 can be transformed into an initial value problem of a second-order equation for x 1 . It is (a) x 00 1 - 6 x 0 1 + 8 x 1 = 0, x 1 (0) = 2, x 0 1 (0) = - 1 (b) x 00 1 - 6 x 0 1 + 8 x 1 = 0, x 1 (0) = 2, x 0 1 (0) = 13 (c) x 00 1 + 7 x 0 1 - 2 x 1 = 0, x 1

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midterm152sol - Part I Answer each of the following 16...

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