Differential Equations Chapter 3, Computing Solutions

Chapter 3, Computing Solutions From the theory, we know that every initial value problem: ay 00 + by 0 + cy = g(t) y(t0 ) = y0 y 0 (t0 ) = v0 has a solution that can be expressed as: y(t) = c1 y1 + c2 y2 + yp where y1 , y2 form a fundamental set of solutions to the homogeneous equation, and yp (t) is the (particular) solution to the nonhomogeneous equation. We first consider the homogeneous ODE: Solving ay 00 + by 0 + cy = 0 Form the associated characteristic equation (built by using y = ert as the ansatz): √ −b ± b2 − 4ac ar2 + br + c = 0 ⇒ r= 2a so that the solutions depend on the discriminant, b2 − 4ac in the following way (yh refers to the solution of the homogeneous equation): • b2 − 4ac > 0 ⇒ two distinct real roots r1 , r2 . The general solution is: yh (t) = c1 er1 t + c2 er2 t If a, b, c > 0 (as in the Spring-Mass model) we can further say that r1 , r2 are negative. We would say that this system is OVERDAMPED. • b2 − 4ac = 0 ⇒ one real root r = −b/2a. Then the general solution is: yh (t) = e−(b/2a)t (C1 + C2 t) If a, b, c > 0 (as in the Spring-Mass model), the exponential term has a negative exponent. In this case (one real root), the system is CRITICALLY DAMPED. • b2 − 4ac < 0 ⇒ two complex conjugate solutions, r = λ ± iµ. Then the solution is: yh (t) = eλt (C1 cos(µt) + C2 sin(µt)) If a, b, c > 0, then λ < 0. In the case of complex roots, the system is said to the UNDERDAMPED. If λ = 0 (this occurs when there is no damping), we get pure periodic motion, with period 2π/µ. Solving y 00 + p(t)y 0 + q(t)y = 0 Given y1 (t), we can solve for a second linearly independent solution to the homogeneous equation, y2 , by one of two methods: • By use of the Wronskian: There are two ways to compute this, R – W (y1 , y2 ) = Ce− p(t) dt (This is from Abel’s Theorem) – W (y1 , y2 ) = y1 y20 − y2 y10 Therefore, these are equal, and y2 is the unknown: y1 y20 − y2 y10 = Ce− R p(t) dt Summarized Example: y 00 + 2t y 0 − Abel’s Theorem: W (y1 , y2 ) = Ce 2 t2 −2 ln(t) = 0, with y1 = t. = C/t2 Wronskian: ty20 − y2 These should be the same: ty20 −y2 = to get that y2 = t−2 . C t2 is a linear first order equation. Solve it and ignore the constant • By Variation of Parameters (the method that the text uses in Section 3.5), where y2 = u2 (t)y1 (t). See Example 3, p. 171 for an example. Solving for the particular solution. Our two methods were: Method of Undetermined Coefficients and Variation of Parameters. • Method of Undetermined Coefficients This method is motivated by the observation that, a linear operator of the form L(y) = ay 00 + by 0 + cy, acting on certain classes of functions, returns the same class. In summary, we have Table 3.6.1, reproduced below: if gi (t) is: Pn (t) Pn (t)eαt Pn (t)eαt sin(µt) or cos(µt) The ansatz ypi is: ts (a0 + a1 t + . . . an tn ) ts eαt (a0 + a1 t + . . . + an tn ) ts eαt ((a0 + a1 t + . . . + an tn ) sin(µt) + (b0 + b1 t + . . . + bn tn ) cos(µt)) The ts term comes from an analysis of the homogeneous part of the solution. That is, multiply by t or t2 so that no term of the ansatz is included as a term of the homogeneous solution. • Variation of Parameters: Given y 00 + p(t)y 0 + q(t)y = g(t), with y1 , y2 solutions to the homogeneous equation, we write the ansatz for the particular solution as: yp = u 1 y1 + u 2 y2 From our analysis, we saw that u1 , u2 were required to solve: u01 y1 + u02 y2 u01 y10 + u02 y20 =0 =0 From which we get the formulas for u01 and u02 : u01 = −y2 g W (y1 , y2 ) u02 = y1 g W (y1 , y2 )