Differential Equations Variation of Parameters (3.6)

Variation of Parameters (3.6) Summary Given y 00 + p(t)y 0 + q(t)y = g(t), we assume yp = u1 y1 + u2 y2 Where y1 , y2 solve the homogeneous DE. Then u1 , u2 satisfy u01 y1 u01 y10 + + u02 y2 u02 y20 0 y2 g(t) y20 ⇒ u01 = W (y1 , y2 ) =0 = g(t) y1 0 0 y1 g(t) u02 = W (y1 , y2 ) Examples 1. Use Variation of Parameters to find the general solution: 4y 00 − 4y 0 − 8y = 8e−t In standard form, y 00 − y 0 − 2y = 2e−t so that r = −1, 2 ⇒ y1 = e2t y2 = e−t g(t) = 2e−t This was the error from class- We need to write the DE in standard form to determine g(t)... This error made an important point! We need W (y1 , y2 ): W = e2t e−t 2e2t −e−t = −3et Substituting into the formulas, e−t 2e−t −e−t −3et 0 u01 = = −2e−2t 2 −3t = e −3et 3 u02 = e2t 0 2e2t 2e−t −3et Therefore, u1 = − 29 e−3t + C1 and u2 (t) = − 32 t + C2 , so that 2 2 yp = − e2t + C1 e2t − te−t + C2 e−t 9 3 From which we observe that the only term that is not part of yh is: 2 yp = − te−t 3 so the full solution is 2 y(t) = C1 e−t + C2 e2y − te−t 3 1 = 2et 2 = − −3et 3 We’ll verify this using the Method of Undetermined Coefficients. The homogeneous part of the equation remains the same, but we guess that yp = Ate−t since Ae−t is part of the homogeneous solution. Substituting this into the DE, we get: −8yp −4yp0 4yp00 8e−t = −8Ate−t = (−4A +4At)e−t = (−8A +4At)e−t = (−12A +0t)e−t ⇒ A=− 2 3 so we get the same solution as before. 2. Use any method to find the general solution: y 00 − 2y 0 + y = et + 3et 1 + t2 SOLUTION: First, the solution to the characteristic equation is r = 1, 1 Now, we will use both Variation of Parameters and Method of Undetermined Coefficients. First, we’ll find the particular solution if g(t) = et /(1 + t2 ), then we’ll find the other one. Here we go for Variation of Parameters: y1 = et Therefore y2 = tet t u01 = e −tet (1+t 2) e2t = g(t) = −t 1 + t2 ⇒ et 1 + t2 W = e2t 1 u1 = − ln(1 + t2 ) 2 t e et (1+t 2) 1 ⇒ u2 = tan−1 (t) 1 + t2 NOTE: We do not need the constants, as this will just bring in components from the homogeneous part of the solution. u02 = e2t = Therefore, the particular part of the solution is: 1 yp1 (t) = − et ln(1 + t2 ) + tet tan−1 (t) 2 Next, we solve the other part. We guess Aet , but we’ll need to multiply by t2 : yp2 = At2 et yp0 2 = (At2 + 2At)et yp002 = (At2 + 4At + 2A)et We substitute into the DE and solve for the coefficients: yp −2yp0 yp00 3et = (At2 )et = (−2At2 −4At )et 2 = (At +4At +2A)et = (0 0 2A)et 2 ⇒A= 3 2 Therefore, the overall solution is: y(t) = e t 3 1 C1 + C2 t + t2 − et ln(1 + t2 ) + tet tan−1 (t) 2 2   3. Use the Variation of Parameters to solve t2 y 00 − t(t + 2)y 0 + (t + 2)y = 2t3 y1 (t) = t y2 (t) = tet SOLUTION: g(t) = 2t and W (y1 , y2 ) = t2 et , so u01 = −2t2 et = −2 t 2 et u02 = 2t2 = 2e−t t2 et ⇒ u1 = −2t Similarly, ⇒ u2 = −2e−t so yp = (−2t)(t) + (−2e−t )(tet ) = −2t2 − 2t ⇒ yp (t) = −2t2 Summary of Techniques 1. We had two techniques for finding a fundamental set of solutions to a homogeneous linear second order differential equation (with possibly non-constant coefficients): • Reduction of Order: Assume y2 = v(t)y1 (t), and substitute into the DE. You’ll get a differential equation in v 00 and v 0 (so first order in v 0 ) that we can then solve. • Wronskian: We can compute the Wronskian in two ways- Abel’s Theorem and the usual method. This gives a first order DE in y2 (given y1 ) that we can solve. The second method is probably easier to use in many instances. 2. We had two techniques for finding the particular solution to a non-homogeneous second order linear DE (with forcing function g(t)): • Method of Undetermined Coefficients (g(t) has to be of a certain type). • Variation of Parameters (This section). The second method is more general than the first, but can be more difficult to implement (because of the integrals). 3