MATH 33BOutline3330spring2007

MATH 33BOutline3330spring2007 - Mathematics 33B This is...

Info icon This preview shows pages 1–3. Sign up to view the full content.

Mathematics 33B This is intended to be an outline for the entire course and a guide for studying for the final exam. I. First Order Equations (Sections 2.1,2.2, 2.4-2.9) We covered quite a bit here, but it really all comes under the following topics: i) Solving y + a ( t ) y = g ( t ) for the unknown function y ( t ) (Section 2.4). ii) Exact equations in general, and y = f ( t ) h ( y ) in particular (Section 2.2 and 2.6). iii) Asymptotic stability and instability of solutions to autonomous equations (Sec- tion 2.9). iv) Learning to use these differential equations in simple applications (Section 2.5, 3.1 and 3.3). v) The dreaded existence and uniqueness theory (Section 2.1 and 2.7). With regard to (2.4) you should definitely know how to explain that y + a ( t ) y = g ( t ) is equivalent to ( e A ( t ) y ) = e A ( t ) g ( t ) , where A ( t ) = integraltext a ( t ) dt is any anti-derivative of a ( t ), as I asked some of you to do on the first hour exam. For y = f ( t ) h ( y ) be on the lookout for values of y where h ( y ) = 0. Remember, if h ( y 0 ) = 0, then the constant function y ( x ) = y 0 will be a solution that you will NOT find by the standard method of reducing the equation to integraltext ( h ( y )) 1 dy = integraltext f ( t ) dt . There are some basic facts from Section 2.6 which you need to know: a) The statement that y ( x ) is a solution to y = f ( x, y ) is equivalent to the statement that the differential form ω = μ ( x, y ) dy μ ( x, y ) f ( x, y ) dx is zero on y = y ( x ). When ω = ( ∂F/∂x ) dx + ( ∂F/∂x ) dy , then the level curves F ( x, y ( x )) = C are solutions to y = f ( x, y ). b) The differential form ω = Pdx + Qdy equals ( ∂F/∂x ) dx + ( ∂F/∂x ) dy for some function F on a rectangle if and only if ∂P/∂y = ∂Q/∂x . When this happens, ω is called “exact”. c) When ω is exact, you can find F in the form F = integraldisplay Pdx + h ( y ) or F = integraldisplay Qdy + g ( x ) , where you have to choose h so that ∂F/∂y = Q , or you have to choose g so that ∂F/∂x = P . On the first hour exam I asked some of you to explain why that works. I might ask again.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

2 d) If ω is not exact, you can always multiply it by an integrating factor μ ( x, y ) to make it exact. Unfortunately, finding μ is usually very difficult. It is not difficult to look for integrating factors which just depend on one variable, and I might ask you to do that. For all first order equations remember that two solutions of the same equation “cannot cross” each other, meaning if two solutions agree at a point, y ( t 1 ) = z ( t 1 ), then they must agree on their whole interval of existence, that means y ( t ) z ( t ). This follows from the uniqueness part of the existence and uniqueness theorem, and it can fail when the hypotheses of that theorem are not satisfied. That happens in examples like y = y 1 / 3 with y ( t 0 ) = 0. The proof of the existence and uniqueness theorem is well beyond the scope of this course (even though the proof of the uniqueness half is posted on the home-page), but you should know what it says .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern