INTEGRATING FACTORS 5 minute review. Remind students how to use integrating factors to solve linear dy first-order differential equations. The example dx +2xy = x is a good one, if needed. Class warm-up. Find the solution to dy 8 + 8x7 y = ex−x , dx satisfying y = 2 when x = 1. Problems. Choose from the below. 1. A basic example. Find the general solution to the differential equation dy 2y sin x + = 2 . dx x x 2. An example with initial conditions. Find the solution of the differential equation dy y exp + =x dx x which satisfies x = 1 at y = 1. 3. Daily temperature cycles. Recall Newton’s law of cooling, which gives a differential equation describing the temperature of a body T in terms of the ambient temperature Ta , namely dT = −k(T − Ta ). dt In the videos we solved this differential equation when Ta was constant. Instead, let’s assume Ta = sin t (a crude approximation of a daily temperature cycle). Find the solution of the resulting differential equation. 4. A non-linear equation. By considering the expansion of (s − t)(s + t − u), find two independent families of solutions to 2 dy dy −x − y 2 + xy = 0. dx dx 1 2 INTEGRATING FACTORS 8 For the warm-up, the integrating factor is ex , and the particular solution is y = 8 8 ex−x + e1−x . Selected answers and hints. 1. It’s y= c − cos x . x2 2. The solution turns out to be 1 1 5 y = x ln x − x + . 2 4 4x 3. The integrating factor is (as in the example considered in the videos) ekt , and the solutions turn out to be k(k sin t − cos t) + ce−kt , T = k2 + 1 where c is a constant. dy dy 4. The equation factorises as dx = 0, and we can find one −y + y − x dx family of solutions to each bracket. This leads to the solutions y = Aex and y = x − 1 + Be−x . For more details, start a thread on the discussion board.
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