Differential Equations Practice: Euler Equations & Regular Singular Points Page 1

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1 Differential Equations Practice: Euler Equations & Regular Singular Points Page 1 Questions Eample (5.4.1) Determine the solution to the differential equation y + 4y + y = 0 that is valid in any interval not Eample (5.4.) Determine the solution to the differential equation ( + 1) y + 3( + 1)y + 3 y = 0 that is valid in any 4 interval not Eample (5.4.9) Determine the solution to the differential equation y 5y + 9y = 0 that is valid in any interval not Eample (5.4.17) Find all the singular points of the differential equation y +(1 )y +y = 0, and determine whether each one is regular or irregular. Eample (5.4.0) Find all the singular points of the differential equation (1 )y + y + 4y = 0, and determine whether each one is regular or irregular. Eample (5.4.1) Find all the singular points of the differential equation (1 ) y + (1 )y + (1 + )y = 0, and determine whether each one is regular or irregular. Eample (5.4.41) For the differential equation y + 3y + y = 0, show that = 0 is a regular singular point. Show that there is only one nonzero solution of the form y = a n n. In general, for regular singular points 0 there may be no solutions of the form a n( 0 ) n. Solutions Eample (5.4.1) Determine the solution to the differential equation y + 4y + y = 0 that is valid in any interval not The is an Euler equation (since the power of in the coefficient is the same as the derivative), and the singular point is = 0. Assume a solution looks like y = r. y = r, y = r r 1, y = r(r 1) r y + 4y + y = 0 r(r 1) r + 4r r + r = 0 r(r 1) + 4r + = 0 This is the indicial equation. Solving, we find (r + )(r + 1) = 0, so the roots are r = and r = 1. The general solution is therefore y() = c c, which is valid for > 0. This is also the solution if < 0, so the general solution is y() = c 1 + c, 0. Eample (5.4.) Determine the solution to the differential equation ( + 1) y + 3( + 1)y + 3 y = 0 that is valid in any 4 interval not The is an Euler equation (since the power of + 1 in the coefficient is the same as the derivative), and the singular point is = 1. Assume a solution looks like y = ( + 1) r.

2 Differential Equations Practice: Euler Equations & Regular Singular Points Page y = ( + 1) r, y = r( + 1) r 1, y = r(r 1)( + 1) r ( + 1) y + 3( + 1)y y = 0 r(r 1)( + 1) r + 3r( + 1) r ( + 1)r = 0 r(r 1) + 3r = 0 This is the indicial equation. Solving, we find (r + 3/)(r + 1/) = 0, so the roots are r = 3/ and r = 1/. The general solution is therefore y() = c 1 ( + 1) 3/ + c ( + 1) 1/, which is valid for > 1. For < 1, the solution is y() = c / + c + 1 1/. For 1, the solution is y() = c / + c + 1 1/. Eample (5.4.9) Determine the solution to the differential equation y 5y + 9y = 0 that is valid in any interval not The is an Euler equation (since the power of in the coefficient is the same as the derivative), and the singular point is = 0. Assume a solution looks like y = r. y = r, y = r r 1, y = r(r 1) r y 5y + 9y = 0 r(r 1) r 5r r + 9 r = 0 r(r 1) 5r + 9 = 0 This is the indicial equation. Solving, we find (r 3) = 0, so the root is r = 3 of multiplicity. Therefore, one solution is y 1 () = 3. If we forget the second solution looks like y () = 3 ln for a repeated root, we can always work it out using reduction of order. Assume y() = v()y 1 () = v 3. y = v 3 + 3v y = v 3 + 6v + 6v

3 Differential Equations Practice: Euler Equations & Regular Singular Points Page 3 Substitute into the differential equation, and determine v: y 5y + 9y = 0 (v 3 + 6v + 6v) 5(v 3 + 3v ) + 9(v 3 ) = 0 v 5 + 6v 4 + 6v 3 5v 4 15v 3 + 9v 3 = 0 v + v = 0 dv d + v = 0 dv d v = ln v + c 1 = ln ln v = ln 1 c 1 v = e c1 1 v = c, c = e c1 c dv = d v = c ln + c 3 Another solution is therefore y() = v()y 1 () = c 3 ln + c 3 3. This is the general solution, and a fundamental set of solutions is y 1 () = 3 and y () = 3 ln, which is valid for 0. Eample (5.4.17) Find all the singular points of the differential equation y +(1 )y +y = 0, and determine whether each one is regular or irregular. Identify p() = 1 and q() = = 1. Since p() is not analytic at = 0 (meaning there is no Taylor series with nonzero radius of convergence about = 0), = 0 is a singular point. Since p() = 1 and q() = are both analytic at = 0, we have = 0 is a regular singular point. Eample (5.4.0) Find all the singular points of the differential equation (1 )y + y + 4y = 0, and determine whether each one is regular or irregular. Identify p() = 3 (1 ) and q() = 4 (1 ). Since p() is not analytic at = 0, 1, 1 (meaning there is no Taylor series with nonzero radius of convergence about = 0, 1, 1), = 0, 1, 1 are singular points. Also, q() is not analytic at these points. We need to classify these points, taking each in turn. = 0: Consider ( 0)p() = (1 ) and ( 4 0) q() = 1. Although q() is analytic at = 0, since p() is not analytic at = 0, the point = 0 is an irregular singular point. = 1: Consider ( 1)p() = 3 (1 + ) and ( 1) q() = singular point. = 1: 4(1 ) 3. Both these are analytic at = 1, so = 1 is a regular (1 + )

4 Differential Equations Practice: Euler Equations & Regular Singular Points Page 4 Consider ( + 1)p() = singular point. 3 (1 ) and ( + 4(1 + ) 1) q() = 3. Both these are analytic at = 1, so = 1 is a regular (1 ) Eample (5.4.1) Find all the singular points of the differential equation (1 ) y + (1 )y + (1 + )y = 0, and determine whether each one is regular or irregular. Identify p() = (1 ) (1 ) = (1 + ) and q() = (1 )(1 + ) (1 ) = 1 (1 ) (1 + ). Since p() is not analytic at = ±1 (meaning there is no Taylor series with nonzero radius of convergence about = ±1), = ±1 are singular points. Also, q() is not analytic at these points. We need to classify these points, taking each in turn. = 1: Consider ( + 1)p() = (1 )(1 + ) and ( + 1) q() = is an irregular singular point. = 1: Consider ( 1)p() = point. ( + 1). Since ( + 1)p() is not analytic at = 1, = 1 (1 ) (1 + ) and ( 1) q() = 1. Both these are analytic at = 1, so = 1 is a regular singular 1 + Eample (5.4.41) For the differential equation y + 3y + y = 0, show that = 0 is a regular singular point. Show that there is only one nonzero solution of the form y = a n n. In general, for regular singular points 0 there may be no solutions of the form a n( 0 ) n. Identify p() = 3 and q() = 1. Since p() is not analytic at = 0, = 0 is a singular point. Consider p() = 3 and q() =. Since both of these are analytic at = 0, = 0 is a regular singular point. Assume: y = y = y = a n n (n + 1)a n+1 n (n + )(n + 1)a n+ n

5 Differential Equations Practice: Euler Equations & Regular Singular Points Page 5 Substitute into the differential equation: (n + )(n + 1)a n+ n + 3 (n + 1)a n+1 n + a n n = 0 (n + )(n + 1)a n+ n (n + 1)a n+1 n + a n n+1 = 0 (n + 1)na n+1 n + 3 (n + 1)a n+1 n + a n 1 n = 0 (n + 1)na n+1 n + 3a (n + 1)a n+1 n + a n 1 n = 0 3a [(n + 1)na n+1 + 3(n + 1)a n+1 + a n 1 ] n = 0 For this to be true for all values of, the coefficients of powers of must be zero. This leads to the relations: 3a 1 = 0 (n + 1)na n+1 + 3(n + 1)a n+1 + a n 1 = 0, n = 1,, 3,... These are the recurrence relations. Solving for the coefficients, we get a n 1 a n+1 =, n = 1,, 3,... (n + 1)(n + 3) a 0 = unspecified, not equal to zero a 1 = 0 a = a 0 5 a 3 = 0 a 4 = a 4 9 = a Finding the pattern here is difficult, and see we only need to show there is only one solution of the form a n n, the important thing to note is that all the odd terms are zero, and the even terms all have the constant a 0 in them. ( The solution we find is y() = a ) So we only get one solution, and we cannot write a general solution. This problem shows the failure of using Taylor series when looking for a series solution about a regular singular point.