18.03 Class 4, Feb 10, 2010 First order linear equations: integrating factors [1] First order homogeneous linear equations [2] Newtonian cooling [3] Integrating factor (IF) [4] Particular solution, transient, initial condition [5] General formula for IF Definition: A "linear ODE" is one that can be put in the "standard form" _______________________________ | | | r(t)x' + p(t)x = q(t) | x = x(t) |_______________________________| r(t), p(t) are the "coefficients" [I may have called q(t) also a coefficient also on Monday; this is not correct, fix it if I did.] The left hand side represents the "system," and the right hand side arises from an "input signal." A solution x(t) is a "system response" or "output signal." We can always divide through by r(t), to get an equation of the Reduced standard form: _______________________________ | | | x' + p(t)x = q(t) | x = x(t) (*) |_______________________________| The equation is "homogEneous" if q is the "null signal," q(t) = 0 . This corresponds to letting the system evolve in isolation: In the bank example, no deposits and no withdrawals. In the RC example, the power source is not providing any voltage increase. The homogeneous linear equation x' + p(t) x = 0 (*)_h is separable. Here's the solution, in general on the left, with an example (with p(t) = 2t ) on the right: x' + p(t)x = 0 x' + 2tx = 0 Separate: dx/x = - p(t) dt dx/x = - 2t dt Integrate: ln|x| = - int p(t) dt + c ln|x| = - t^2 + c Exponentiate: |x| = e^c e^{ - int p(t) dt } |x| = e^c e^{-t^2} Eliminate the absolute value and reintroduce the lost solution:x = C e^{- int p(t) dt} x = C e^{-t^2} In the example, we chose a particular anti-derivative of k , namely kt. That is what I really have in mind to do in general. The constant of integration is taken care of by the constant C . So the general solution to (*)_h has the form C x_h , where x_h is *any* nonzero solution: x_h = e^{- int p(t) dt} , x = C x_h We will see that the general case can be solved by an algebraic trick that produces a sequence of two integrations. [2] Example: Diffusion, e.g. of heat. About this time of year I start to think about summer. I put my rootbeer in a cooler but it still gets warm. Let's model its temperature by an ODE. x(t) = root beer temperature at time t . The greater the temperature difference between inside and outside, the faster x(t) changes. Simplest ("linear") model of this: x'(t) = k ( T_ext(t) - x(t) ) where T_ext(t) is the "external" temperature. Sanity check: When T_ext(t) > T(t), x'(t) > 0 (assuming k > 0 ). We get a linear equation: x' - k x = k T_ext This is "Newton's law of cooling." k could depend upon t and we would still have a linear equation, but let's suppose that we are not watching the process for so long that the insulation of the cooler starts to break down! Systems and signals analysis: The system is the cooler. The output signal = system response is x(t) , the temperature in the cooler. The input signal is the external temperature T_ext(t) . Note that the right-hand side is k times the input signal, not the input signal itself. What constitutes the input and output signals is a matter of the interpretation of the equation, not of the equation itself. Question 4.1: k large means 1. good insulation 2. bad insulation Blank. don't know.k is small when the insulation is good, large when it is bad. It's zero when the insulation is perfect. k is a COUPLING CONSTANT When it is zero, the temperature inside the cooler is decoupled from the temperature outside. In the construction industry, a number like k is pasted on windows; it's called the U-value of the window. Let's take k = 1/3 , for example. Suppose the temperature outside is rising at a constant rate: say T_ext = 60 + 6t (in hours after 10:00) and we need an initial condition: let's say x(0) = 32 . So the IVP is x' + (1/3) x = 20 + 2t , x(0) = 32 . (cooler) This isn't separable: it's something new. We'll describe a method which works for ANY first order linear ODE. [3] Method: Integrating factors (Euler) This method is based on the product rule for differentiation: (d/dt) ( u x ) = ux' + u'x For example, suppose we have the equation t x' + 2 x = t (This is not separable; it is linear and in standard form, but not reduced standard form.) Here's a *trick*. Multiply both sides by t : t^2 x' + 2t x = t^2 The left hand side is now the derivative of a product: (d/dt) (t^2 x) = t^2 We can solve this by integrating: t^2 x = t^3/3 + c so x = t/3 + c t^{-2} [In the first lecture, I posed this (with a different righthand side) as a flashcard problem, but I did it just after describing the calculation of an integrating factor for a *reduced* equation. The reduced equation is x' + 2x/t = 1 , and this has integrating factor t^2 . So it was a poorly placed question.] That was great! The factor t we multiplied by is an "integrating factor." I guessed it here. Often you can. The factor to use in the cooler equation and other equations may not be so obvious. Here's a calculation, for a linear equation in reduced form,x' + p(t)x = q(t) Multiply both sides by u u x' + p u x = u q In order for the right hand side to be (d/dt)(ux) = ux' + u'x, the function u must satisfy the differential equation u' = p u This is separable, and we'll carry out the separation in general in a minute. In the cooler equation, the coefficient p(t) is constant. In that case we have the natural growth equation! u = e^{pt} (I am choosing a value for the constant of integration, because I need just one integrating factor, any one.) In the case of the cooler problem, p = 1/3 , so we have: (d/dt) (e^{t/3} x) = (20 + 2t) e^{t/3} Integrate: e^{t/3} x = 60 e^{t/3} + \int 2t e^{t/3} dt Um. Parts: \int u dv = uv - \int v du u = 2t , dv = e^{t/3} dt du = 2dt , v = 3 e^{t/3} e^{t/3} x = 60 e^{t/3} + 6 t e^{t/3} -
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