This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Exam I This is not the original form, but the questions are the same. 1. Solve the initialvalue problem: y + y = 1 + e x , y (0) = 5 . We use the method of integrating factors. Recall = exp R p ( x ) dx , so here = e x . Multiplying both sides by we have e x y + e x y = e x + e 2 x . We recognize the lefthand side as d dx ( e x y ). Thus d dx ( e x y ) = e x + e 2 x . Integrating both sides, we find e x y = e x + 1 2 e 2 x + C. (1) We plug in our initial condition and find 5 = 1 + 1 2 + C, so C = 7 2 . Plugging this into (1) and dividing both sides by e x , we conclude y = 1 + 1 2 e x + 7 2 e x . 2. Find the general solutions of dy dx + y x = g ( x ) where g ( x ) = , for x < 1 1 , for x 1 . We solve problems like this in two steps: For x < 1: Here dy/dx + y/x = 0. Using an integrating factor of = x , we see d dx ( xy ) = 0 and hence y = C/x here. For x 1: Here dy/dx + y/x = 1. Again, = x , so now d dx ( xy ) = x . Thus xy = x 2 2 + C 2 , so y = x 2 + C 2 x ....
View
Full
Document
 Fall '07
 COSTIN
 Differential Equations, Equations, Factors

Click to edit the document details