Section 2.1 (First Order) Linear DEs; Method of Integrating Factors. General first order linear DEs Standard Form; y'(t) + p(t) y = g(t)

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1 Section 2.1 (First Order) Linear DEs; Method of Integrating Factors Key Terms/Ideas: General first order linear DEs Standard Form; y'(t) + p(t) y = g(t) Integrating factor; a function μ(t) that transforms the DE into an equivalent DE that is easier to solve Integration techniques required Finding a critical initial value that separates solutions that behave one way from others that behave quite differently

2 The general form of a first order DE is If we can rearrange the DE into the form y'(t) + p(t)y = g(t) then we say that the DE is first order linear in y. Note that functions p(t) and g(t) are dependent only on t. There are times when we must algebraically rearrange the DE to see if it is first order linear and obtain the standard form y'(t) + p(t)y = g(t). Examples: DE Standard Form p g y' = 5y+3 y'- 5y = y' + 8y = t 4 xy'-x 2 y = e x -x t 2 sin(t) cos(t)y = sin(t)y'

3 Example: Is DE t 2 y' + sin(t) y 3 = ln(t) first order linear? Explain. How do we solve first order linear ODEs? Let's look at the form of the equation, in particular the left side of y'(t) + p(t)y = g(t). The left side, y'(t) + p(t)y looks like a "piece of a product rule". Recall that d y (t) (t)y' '(t)y dt y' p(t)y left side y'(t) + p(t)y = g(t) Strategy: A way to proceed is to try to find a function (t) so that when we multiply both sides of the DE y'(t) + p(t)y = g(t) by (t) the left side becomes the exact derivative of the product (t) y. Such a function (t) is called an integrating factor.

4 Let s apply this strategy; the math requires some algebra that leads to a DE to solve for function (t). We want Now expand the left side: We have y'(t) = - p(t)y + g(t) so replace y'(t) to get Set equal to Expand by multiplying and simplify. when simplified gives Factor. This will be true when Thus we have a DE to solve to find function (t), the integrating factor.

5 Solving this DE for (t) by separating the variables we get Assuming we can compute we have found the function so if we multiply both sides of the DE y'(t) + p(t)y = g(t) by (t) then it can be immediately written in the form If we can compute the function (t) it is an integrating factor of the DE.

6 Here is how we use the integrating factor to solve the first order linear DE y'(t) + p(t)y = g(t) as follows. (Note it must be in standard form!) Thus we have a formula for the solution of a first order linear DE. BUT it requires that you be able to compute TWO integrals:

7 Example: Solve General solution of the DE.

8 Example: Solve This is in standard form so p(t) = -2 and g(t) = 4 t. Need integration by parts. We get Now solve for y: Next we look at the direction field for the DE and some of the integral curves.

9 Direction field for DE

10 Note that for different choices of an initial condition the resulting solution of the IVP can behave differently.

11 Another Example: Note the solution doesn t exist for t = 0. What is the solution when C = 0? Describe it geometrically. If we look at a sketch of integral curves we see different behaviors. We next focus on integral curves for t >0. t y (1,2) t 1 2

12 The IVP solution curve. (0,0) The solution of the IVP is labeled with the initial condition. Note that it becomes unbounded and is asymptotic to the positive y-axis as t 0 from the right. This is the effect of the infinite discontinuity in the coefficient p(t) at the origin. The function y = t 2 + (1/t 2 ) for t < 0 is not part of the solution of this initial value problem.

13 This is the first example in which the solution fails to exist for some values of t. Again, this is due to the infinite discontinuity in p(t) at t =0, which restricts the solution to the interval 0< t <. From the picture of the integral curves we see that if the initial condition is specified at t = 1 then different y-coordinates can produce solutions which behave quite differently as t 0 +. (Some go to + and others go to -.). To determine a value for the y-coordinate that acts as a critical value acts to separate the behaviors consider a general initial condition at t = 1 given by (0,0) t-axis

14 Returning to the general solution y = t 2 + C/t 2 we apply initial condition y(1) = y 0. We find that C = y 0-1 so we have 2 y0-1 y(t) = t + We consider t > 0 and y 0 1. t 2 Now let s consider the limit of y(t) as t 0 from the right for values of y 0 >1 and then values of y 0 <1. Case y 0 > 1: Case y 0 < 1: So the behavior as t 0 from the right is separated by the integral curve that has initial condition y(1) = 0.

15 Here the diagram shows the phenomena known as a bifurcation. A bifurcation occurs when a small (smooth) change made to the parameter values (here the initial condition) of an ODE causes a sudden 'qualitative' change in its behavior. In this case the behavior of the integral curves drastically changes for the initial condition y(1) = y 0 for the cases y 0 <1 and y 0 >1. Geometrically the integral curve y = t 2 separates the behaviors. Ref: