This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ( x ) exp Z x p ( t ) dt , which gives solution y ( x ) = 1 ( x ) Z x ( t ) f ( t ) dt + C , with C an arbitrary constant. In this problem, we can identify p ( t ) = cos t from which = 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 = 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 = 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....
View Full
Document
 Winter '09
 NilesA.Pierce

Click to edit the document details