230 likes | 549 Views
Engineering Mathematics Class #7 Second-Order Linear ODEs ( Part3). Sheng-Fang Huang. 2.5 Euler–Cauchy Equations. Euler–Cauchy equations are ODEs of the form (1) x 2 y " + axy ' + by = 0 a and b are constants, and y ( x ) is unknown. Let (2) y = x m
E N D
Engineering Mathematics Class #7Second-Order Linear ODEs (Part3) Sheng-Fang Huang
2.5 Euler–Cauchy Equations • Euler–Cauchy equations are ODEs of the form (1) x2y" + axy' + by = 0 • a and b are constants,and y(x) is unknown. Let (2)y =xm Therefore, substitute y' = mxm-1 and y'' = m(m – 1)xm-2 into (1). Then, x2m(m – 1)xm-2 + axmxm-1 + bxm = 0.
We now see that (2) was a rather natural choice because we have obtained a common factor xm. Dropping it, we have the auxiliary equation m(m – 1) + am +b = 0 or (3) m2 + (a – 1)m +b = 0. (Note: a – 1, not a.) Hence y =xmis a solution of (1) if and only if m is a root of (3).
Case I: Different Real Roots • If the roots m1 and m2 are real and different, then solutions are and They are linearly independent since their quotient is not constant (basis). • The corresponding general solution for all these x is (5) (c1, c2 arbitrary).
Example 1Different Real Roots • The Euler–Cauchy equation x2y" + 1.5xy' – 0.5y = 0 has the auxiliary equation m2 + 0.5m – 0.5 = 0. • The roots are 0.5 and –1. Hence a basis of solutions for all positive x is y1 = x0.5 and y2 = 1/x and gives the general solution (x > 0).
Case II: Double Roots • m = 1/2(1 – a) iff. (1 – a)2 – 4b = 0. That is, (6) Find the basis:
Example 2:Double Root • The Euler–Cauchy equation x2y" – 5xy' + 9y = 0 has the auxiliary equation m2 – 6m + 9 = 0. It has the double root m = 3, so that a general solution for all positive x is y = (c1 + c2 ln x) x3.
Case III. Complex Conjugate Roots • The case of complex roots is of minor practical importance, and it suffices to present an example that explains the derivation of real solutions from complex ones.
2.7 Nonhomogeneous ODEs • In this section we proceed from homogeneous to nonhomogeneous linear ODEs (1) y" + p(x)y' + q(x)y =r(x) where r(x) ≠ 0. We shall see that a “general solution” of (1) is the sum of a general solution of the corresponding homogeneous ODE (2) y" + p(x)y' + q(x)y = 0 and a “particular solution” of (1).
DEFINITION General Solution, Particular Solution • A general solution of the nonhomogeneous ODE (1) on an open interval I is a solution of the form • (3) y(x) = yh(x) + yp(x); • yh = c1y1 + c2y2 is a general solution of the homogeneous ODE on I. • yp is any solution of (1) on I containing no arbitrary constants. • A particular solution of (1) on I is a solution obtained from (3) by assigning specific values to c1 and c2 in yh.
Method of Undetermined Coefficients • The method of undetermined coefficients is suitable for linear ODEs with constant coefficients a and b (4) y" + ay' + by =r(x) Table 2.1 shows the choice of ypfor practically important forms of r(x). Corresponding rules are as follows.
Choice Rules for the Method of Undetermined Coefficients (a) Basic Rule. • Choose yp from Table 2.1 and determine its undetermined coefficients by substituting yp and its derivatives into (4). (b) Modification Rule. • If a term in your choice for yp happens to be a solution of the homogeneous ODE, multiply your choice of yp by x (or by x2 if this solution corresponds to a double root of the characteristic equation of the homogeneous ODE).
Choice Rules for the Method of Undetermined Coefficients (c) Sum Rule. • If r(x) is a sum of functions in the first column of Table 2.1, choose for yp the sum of the functions in the corresponding lines of the second column. • The Sum Rule follows by noting that the sum of two solutions of (1) with r =r1 and r =r2 (and the same left side!) is a solution of (1) with r =r1 + r2.
Example1: Basic Rule (a) • Solve the initial value problem (5) y" + y = 0.001x2, y(0) = 0, y'(0) = 1.5. • Solution.
Example 2:Modification Rule (b) • Solve the initial value problem (6) y" + 3y' + 2.25y = –10 e-1.5x, y(0) = 1, y'(0) = 0. • Solution.
Example 3: Sum Rule (c) • Solve the initial value problem (7) y" + 2y' + 5y =e0.5x + 40 cos 10x – 190 sin 10x, y(0) = 0.16, y'(0) = 40.08. • Solution.
Example 3: Sum Rule (c) Solution.