Ordinary Differential Equations (MA102 Mathematics II) Shyamashree Upadhyay IIT Guwahati Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 1 / 13
Variation of parameters Let us first see a procedure (called VARIATION OF PARAMETERS procedure) to find a particular solution of a linear first-order DE dy + P(x)y = f (x) (1) dx on an interval I om which the functions P(x) and f (x) are continuous. PLEASE LOOK AT THE BOARD. This procedure is applicable to linear higher-order equations as well. To adapt the method of variation of parameters to linear second-order DE a 2 (x)y + a 1 (x)y + a 0 (x)y = g(x) (2) we begin as follows we put the DE in standrad form y + P(x)y + Q(x)y = f (x) (3) by dividing through by the lead coefficient a 2 (x). Here we assume that P(x), Q(x), f (x) are continuous on some interval I. equation (3) is an analog of (1). There is no difficulty in obtaining the complementary function y c of (2) when the coefficients are constants. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 2 / 13
Variation of Parameters contd... The Assumptions: Analogous to the assumption y p = u(x)y 1 (x) that we used earlier to find a particular solution y p of the linear first-order equation (1), for the linear second-order equation (2) we seek a solution of the form y p = u 1 (x)y 1 (x) + u 2 (x)y 2 (x) (4) where y 1 and y 2 form a fundamental set of solutions on I of the associated homogeneous form of (2). Using the product rule to differentiate y p twice, we get y p = u 1 y 1 + y 1u 1 + u 2y 2 + y 2u 2 y p = u 1 y 1 + 2u 1 y 1 + y 1 u 1 +u 2 y 2 + 2u 2 y 2 + y 2 u 2 Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 3 / 13
Variation of Parameters contd... Substituting (4) and the foregoing derivatives into (3) and grouping terms yields y p + P(x)y p + Q(x)y p = d dx [y 1u 1 + y 2u 2 ] +P(x)[y 1 u 1 + y 2u 2 ] + y 1 u 1 + y 2 u 2 = f (x) (5) Because we seek to determine two unknown functions u 1 and u 2, reason indicates that we need two equations. We can obtain these two equations by making the further assumption that the functions u 1 and u 2 satisfy y 1 u 1 + y 2u 2 = 0. This assumption does not come out of the blue but is prompted by the first two terms in (5) since, if we demand that y 1 u 1 + y 2u 2 = 0, then (5) reduces to y 1 u 1 + y 2 u 2 = f (x). We now have our desired two equations, that is two equations for determining the derivaives u 1 and u 2. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 4 / 13
Variation of Parameters contd... By cramer s rule, the solution of the system y 1 u 1 + y 2u 2 = 0 y 1 u 1 + y 2 u 2 = f (x) can be expressed in terms of determinants: u 1 = W 1 W and u 2 = W 2 (6) W where W = y 1 y 2 y 1 y, W 1 = 0 y 2 2 f (x) y and W 2 = y 1 0 2 y 1 f (x). The functions u 1 and u 2 can be found by integrating the results in (6). The determinant W is recognized as the Wronskian of y 1 and y 2. By linear independence of y 1 and y 2 on I, we know that W(y 1 (x), y 2 (x)) 0 for every x in the interval I. Then a particular solution is y p = u 1 y 1 + u 2 y 2 and the general solution is y = y c + y p. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 5 / 13
Example Solve y 4y + 4y = (x + 1)e 2x. It is easy to see that the complementary function y c = c 1 e 2x + c 2 xe 2x. With the identifications y 1 = e 2x and y 2 = xe 2x, we get that the Wronskian W(e 2x, xe 2x ) = e 4x. Since the given DE is already in standard form, we identify f (x) = (x + 1)e 2x. Then we can obtain the determinants W 1 and W 2 using the formulas given in the previous slide. we obtain: W 1 = (x + 1)xe 4x and W 2 = (x + 1)e 4x and so from (6), we get u 1 = (x + 1)xe4x e 4x = x 2 x, u 2 = (x + 1)e4x e 4x = x + 1. Integrating, we get u 1 = x3 3 x2 2 and u 2 = x2 2 + x. Hence y p = ( x3 3 x2 2 )e2x + ( x2 2 + x)xe2x = ( x3 6 + x2 2 )e2x. and y = y c + y p = c 1 e 2x + c 2 xe 2x + ( x3 6 + x2 2 )e2x. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 6 / 13
A Remark Remark (on Constants of Integration): When computing the indefinite integrals of u 1 and u 2, we need not introduce any constants. This is because y = y c + y p = c 1 y 1 + c 2 y 2 + (u 1 + a 1 )y 1 + (u 2 + b 1 )y 2 = (c 1 + a 1 )y 1 + (c 2 + b 1 )y 2 + u 1 y 1 + u 2 y 2 = C 1 y 1 + C 2 y 2 + u 1 y 1 + u 2 y 2. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 7 / 13
Higher-order Equations The method we have just employed for non-homogeneous second-order DE can be generalized to linear n-th order equations that have been put into the standard form y (n) + P n 1 (x)y (n 1) + + P 1 (x)y + P 0 (x)y = f (x) (9) If y c = c 1 y 1 + c 2 y 2 + + c n y n is the complementary solution for (9), then a particular solution is y p = u 1 (x)y 1 (x) + u 2 (x)y 2 (x) + + u n (x)y n (x) where the u k, k = 1, 2,..., n are determined by the n equations. y 1 u 1 + y 2u 2 + + y nu n = 0 y 1 u 1 + y 2 u 2 + + y nu n = 0 y (n 1) 1 u 1 + y(n 1) 2 u 2 + + y(n 1) n u n = f (x). Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 8 / 13
Higher-order Equations contd... The first n 1 many equations in the above system, like y 1 u 1 + y 2u 2 = 0 in (5), are assumptions made to simplify the resulting equation after y p = u 1 (x)y 1 (x) + + u n (x)y n (x) is substituted in (9). In this case, Cramer s rule gives u k = W k, k = 1, 2,..., n W where W is the Wronskian of y 1, y 2,..., y n and W k is the determinat obtained by replacing 0 0 the k-th column of the Wronskian W by the column vector.. When n = 2, we get 0 f (x) (6). Exercise: Solve y 2y y + 2y = e 3x. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 9 / 13
Cauchy-Euler Equation Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 10 / 13
Method of Solution Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 11 / 13
Method of Solution contd... Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 12 / 13
Example Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 13 / 13