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Engineering Mathematics Class #7 Second-Order Linear ODEs ( Part3)

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

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Engineering Mathematics Class #7 Second-Order Linear ODEs ( Part3)

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  1. Engineering Mathematics Class #7Second-Order Linear ODEs (Part3) Sheng-Fang Huang

  2. 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.

  3. 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).

  4. 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).

  5. 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).

  6. Case II: Double Roots • m = 1/2(1 – a) iff. (1 – a)2 – 4b = 0. That is, (6) Find the basis:

  7. 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.

  8. 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.

  9. 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).

  10. 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.

  11. 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.

  12. Table2.1Method of Undetermined Coefficients

  13. 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).

  14. 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.

  15. Example1: Basic Rule (a) • Solve the initial value problem (5) y" + y = 0.001x2, y(0) = 0, y'(0) = 1.5. • Solution.

  16. 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.

  17. 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.

  18. Example 3: Sum Rule (c) Solution.

  19. Fig.51.Solution in Example 3

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