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Assignment Solutions - MATH-315 2010 (Ordinary...

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Unformatted text preview: MATH-315 2010 (Ordinary Differential Equations) Solutions to Written-Assignment #1 February 6, 2010 Solution of 1(a): One can rewrite the given equation as y (y 2 − 1)−4/3 dy = xdx, which is a separable equation, so that a simple integration yields (y 2 − 1)−1/3 = −1/3(x2 + c). The initial condition determines c as c = −3(b2 − 1)−1/3 . Therefore ￿ ￿ ￿−3 12 2 − 1)−1/3 y = 1 + − x + (b . 3 Note that we have to choose the + sign since b > 0. In order to find the interval of definition, we need to know for which values of x, the expression under the radical is non-negative. That is, when is ￿ ￿−3 12 2 −1/3 − x + (b − 1) ≥ −1? 3 To this end, we first make the following simple observation. For A ∈ R − {0}, A−3 ≥ −1 ⇐⇒ A > 0 or A ≤ −1. Now we consider two cases: ￿ ￿ (I) b > 1. Then obviously b2 − 1 > 0. With A = − 1 x2 + (b2 − 1)−1/3 , 3 note that ￿ 3 A > 0 ⇐⇒ |x| < ; 2 − 1)1/3 (b and that A ≤ −1 ⇐⇒ |x| ≥ 1 ￿ 3+ (b2 3 . − 1)1/3 Therefore, in this case, the interval of definition is ￿ ￿ ￿ ￿￿ ￿ 3 3 x ∈ R : |x| < ∪ x ∈ R : |x| ≥ 3 + 2 . (b2 − 1)1/3 (b − 1)1/3 (II) 0 < b < 1. Then obviously, 0 < 1 − b2 < 1. We can write ￿ ￿ 1 A = − x2 − |b2 − 1|−1/3 . 3 In this case, A > 0 cannot happen(why?). ￿ ￿ 1 1 A ≤ −1 ⇐⇒ − x2 − |b2 − 1|−1/3 ≤ −1, ⇐⇒ x2 − 1 ≥ −|b2 − 1|−1/3 , 3 3 ⇐⇒ 12 x ≥ 1 − |b2 − 1|−1/3 . 3 This implies that |x| ≥ ￿ 3 − 3|b2 − 1|−1/3 . Solution of 1(b): This is a Bernoulli equation. So, we use the change of variable u = y −2 to get 1 3 13 − u− 2 u￿ = u− 2 + e2x u− 2 2 or u￿ + 2u = −2e2x . This is a first order linear differential equation with the general solution 1 u = − e2x + ce−2x . 2 Therefore, the general solution for the original equation is seen to be given by y2 = Solution of 1(c): With u = u+x y x − 1 e2x 2 1 . + ce−2x we arrive at du 1 =u+ dx u or This leads to the general solution 12 u = ln |x| + c 2 2 udu = dx . x or y 2 = 2x2 (ln |x| + c). By imposing the initial condition y (1) = −4 the constant c is found to be 8. Thus the explicit solution turns out to be ￿ y = − 2x2 (ln |x| + 8). Solution of 1(d): Using the change of variables x = X + we get the following homogeneous equation 29 7 and y = Y − 38 7 dY 2X + 3Y = . dX 5X + 4Y Next we put U = Y X to get U +X or dU 3U + 2 = dX 4U + 5 4U + 5 1 dU = − dX 4U 2 + 2U − 2 X ￿ ￿ −1/6 14/3 1 + dU = − dX. U + 1 4U − 2 X or By integrating we find the implicit solution −1/6 ln |U + 1| + 7/6 ln |4U − 2| = − ln |X | + c. In terms of x and y , the general solution becomes ￿ ￿ ￿ ￿ ￿ y + 38/7 ￿ ￿ ￿ ￿ ￿ + 7/6 ln ￿4 y + 38/7 − 2￿ = − ln |x − 29/7| + c. −1/6 ln ￿ + 1￿ ￿ x − 29/7 ￿ x − 29/7 Solution of 1(e): First suppose that such integrating factor µ = p(x + y ) exists. We thus have ∂ (M µ) ∂ (N µ) = =⇒ ∂y ∂x ￿ ￿ ∂M ∂N − p(x + y ) = −(M − N )p￿ (x + y ) =⇒ ∂y ∂x ∂M ∂y which means that is a function of x + y only. − ∂N ∂x M −N =− ∂M ∂y − p￿ (x + y ) , p(x + y ) ∂N ∂x M −N 3 Conversely, assume now that ∂M ∂y − ∂N ∂x M −N = q (x + y ), for some one-variable p￿ (t) function q (t). Take p(t) to satisfy q (t) = − , and set p(t) µ = p(x + y ). It now easily follows, from the computations above, that µ is an integrating factor for the original equation. As an application, we consider the situation where M = x2 + 2xy − y 2 , Since ∂M ∂y − ∂N ∂x M −N = N = y 2 + 2xy − x2 . 2 = q (x + y ) x+y is a function of x + y only, where q (t) = 2 , we can choose µ = (x + y )−2 (since t ￿ q (t) = − p ((tt)) with p(t) = t−2 .) By multiplying µ to the equation, we arrive at p the exact equation x2 + 2xy − y 2 y 2 + 2xy − x2 dx + dy = 0, 2 (x + y ) (x + y )2 which has, by integrating, the following implicit solution F (x, y ) = x2 + y 2 = c. x+y Solution of 2: Let t, D(t) and X (t) denote, respectively, the time (hour), the depth of snow at t (unknown unit), and the distance (miles) cleared by the snowplow. We have the following information X (0) = 0, X (3) = 2, X (5) = 3. We assume that the snow started to come t0 hours before 8:00 am. The depth D(t) is proportional to the time (t − t0 ). On the other hand, since the snow was coming at a constant rate k , and since the snowplow was clearing the road at a rate which was inversely proportional to the depth of snow, we have dX k￿ k = = , dt D(t) t − t0 where k ￿ , k are some constants. By solving this simple differential equation we find that X (t) = k ln |t − t0 | + C, 4 for some constant C which is to be determined. Using information above, we arrive at 0 = k ln |t0 | + C, 2 = k ln |3 − t0 | + C, 3 = k ln |5 − t0 | + C. By subtraction, we obtain ￿ ￿ ￿ 3 − t0 ￿ ￿ 2 = k ln ￿ ￿ t0 ￿ , ￿ ￿ ￿ 5 − t0 ￿ ￿ 1 = k ln ￿ ￿ 3 − t0 ￿ . Now by division, we get the following equation in which only t0 is involved: (3 − t0 )3 = t0 (5 − t0 )2 or (t0 − 3)3 = t0 (5 − t0 )2 . (Note that t0 < 0.) The first equation leads to nothing, whereas the second one gets us to the situation where t2 + 2t0 − 27 = 0. 0 Since t0 < 0, we have to dispose the positive solution of this equation and therefore, the only acceptable value for t0 is √ t0 = −1 − 28 ≈ −6.29 which means that the snow started five six and seventeen minutes before eight o’clock in the morning, that is, it started at 1:43 am. Solution of 3: (A). From θ = θ1 − θ2 , we derive tan θ sin θ1 cos θ2 − cos θ1 sin θ2 cos θ1 cos θ2 + sin θ1 sin θ2 = tan(θ1 − tan θ2 ) = = tan θ1 − tan θ2 m1 − m2 = 1 + tan θ1 tan θ2 1 + m1 m2 (B). The slope of the given curve at the point P : (x, y ) is derived as 2x + 2yy ￿ = 0, and at the point P , x m1 = y ￿ (x) = − , y c2 = (x2 + y 2 ). Hence the orthogonal trajectory must satisfy the geometric conditions: x x tan(π /4)(1 + m1 m2 ) = (1 − m2 ) = m2 + y y Therefore, we derive the governing equation: x￿ x￿ 1 − y￿ = y￿ + . y y 5 or (x + y ) y ￿ + x − y = 0. This is a homogeneous equation. To solve this equation, let u(x) = y (x)/x. Hence, we have u−1 xu￿ + u = , u+1 or u+1 dx du = − 2+1 u x ￿ ￿ u+1 1 dx du = tan−1 u + ln(1 + u2 ) = − = − ln |x| + C 2+1 u 2 x It follows that tan−1 or ￿y￿ x + 1 1 ln(x2 + y 2 ) − ln x2 = − ln |x| + C 2 2 tan−1 In polar coordinates, ￿y￿ x + 1 ln(x2 + y 2 ) = C. 2 θ + ln r = C, 6 r = C1 e−θ . ...
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