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Unformatted text preview: ACM 95b/100b Problem Set 1 Solutions Prepared by: Lei Zhang January 7, 2007 Total: 125 points Include grading section: 2 points Problem 1 (10 points) Find the general solution and sketch some integral curves for the following ODE. What is the behavior of the solution as t → ∞ ? y + y cos t = sin t cos t Solution 1 For an equation of the form y ( x ) + p ( x ) y ( x ) = f ( x ) , (1) we define μ ( x ) ≡ exp Z x p ( t ) dt , which gives solution y ( x ) = 1 μ ( x ) Z x μ ( t ) f ( t ) dt + C , with C an arbitrary constant. In this problem, we can identify p ( t ) = cos t from which μ = exp (sin( t )) . Thus, y ( t ) = exp ( sin( t )) Z t exp(sin( s )) sin( s ) cos( s ) ds + C = sin( t ) 1 + C exp( sin( t )) . As t → ∞ , y is bounded, but has no limit; integral curves are given in Figure 1 . Problem 2 (10 points) Inspect the following linear initial value problems and identify the largest open interval on which a unique solution can be guaranteed to exist. Find the solution and comment on the interval over which it exists. ty + 2 y = t 2 t + 1 y (1) = 1 / 2 142 2 4 6 8105 5 Figure 1: Integral curves for the equation in Problem 1b). Solution 2 We will solve the problem using the method of integrating factors. Rewriting this equation so we can use integrating factors, we have y + 2 t y = t 1 + 1 t . So p ( t ) = 2 /t and h ( t ) = t 1 + 1 /t . Both p and h have singularities at t = 0 so the solution is unique on < t < ∞ . The integrating factor is μ = t 2 . Using the formula for y , we have y ( t ) = 1 t 2 Z t s 2 s 1 + 1 s ds + C = 1 4 t 2 1 3 t + 1 2 + C t 2 . We can determine C from the initial condition y (1) = 1 2 = 1 4 1 3 + 1 2 + C, which implies C = 1 12 . Thus, the solution to the IVP is y ( t ) = 1 4 t 2 1 3 t + 1 2 + 1 12 t 2 . This solution is unbounded when t → ∞ , and is defined for < t < ∞ . 2 Problem 3 (10 points) Obtain a continuous solution to the following linear initial value problem with a discontinuous coefficient y + 2 y = g ( x ) , y (0) = 0 , g ( x ) = 1 , ≤ x ≤ 1 , , x > 1 . What is the behavior of y at the discontinuity in g ( x )? Solution 3 For 0 ≤ x ≤ 1, we are solving the equation y + 2 y = 1 subject to the constraint y (0) = 0. We can use the method of integrating factors. We identify p ( x ) = 2, so μ = e 2 x . Using the formula for y we obtain y ( x ) = e 2 x Z x e 2 s ds + C = 1 2 + Ce 2 x . Applying the initial condition, we have y (0) = 0 = 1 2 + C, which implies that C = 1 2 and y ( x ) = 1 2 ( 1 e 2 x ) , ≤ x ≤ 1 . For x > 1, we know that y satisfies the equation y + 2 y = 0 , but it is not immediately obvious what to use for an initial condition. However, y must be continuous at x = 1, so therefore it is appropriate to use y (1) found from our formula above as the initial condition. This is justified as follows: we know that the right hand side of the equation y + 2 y = g ( x ) has a step discontinuity at x = 1, so the left hand side must also have a step discontinuity. If= 1, so the left hand side must also have a step discontinuity....
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 Winter '09
 NilesA.Pierce
 Derivative, Boundary value problem, LHS, RHS

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