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1 M ath E S 1 W inter Last Update: March 1 2, S eries S olutions of Differential Equations Disclaimer: This lecture note tries to provide an alternative approach to the material in Sections in the textbook. It is not a replacement of the textbook. Table of contents Series Solutions of Differential Equations Power series solutions An example Power series and analytic functions More examples Ordinary and singular points Equations with analytic coefficients Cauchy-Euler equations Power series solutions An example. So far we can effectively solve linear equations homogeneous and non-homongeneous) with constant coefficients, but for equations with variable coefficients only special cases are discussed 1 st order, etc. ). Now we turn to this latter case and try to find a general method. The idea is to assume that the unknown function y can be expanded into a power series: We try to determine the coefficients a 0, a 1, Example 1. Solve Solution. Substitute y x) = a 0 + a 1 x + a 2 x ) y 2 x y = 0. 2) y x) = a 0 + a 1 x + a 2 x 2 + 3) into the equation. We have a1 + 2 a 2 x + 3 a 3 x 2 + ) ) 2 a0 x + a 1 x 2 + a 2 x 3 + = 0. 4) Rewrite it to a a 2 2 a 0 ) x + 3 a 3 2 a 1 ) x 2 + = 0. 5) Naturally we require the coefficients to each power of x to be 0: We conclude a 1 = 0 6) 2 a 2 2 a 0 = 0 7) 3 a 3 2 a 1 = 0 8) a 1 = 0, a 2 = a 0, a 3 = 0, 9) We see that the coefficients can be determined one by one. However, as there are infinitely many a i s, we need a general formula. To do this, we return to and write it as y x) = a 0 + a 1 x + a 2 x ) y x) = a n x n. 1 1 )

2 Put this into the equation we have y 2 x y = 0 1 2) ) ) 0 = a n x n 2 x a n x n = n a n ) x n 1 n= 1 2 a n x n+ 1 = n + 1 ) a n+ 1 ) x n n= 1 2 a n 1 x n = a 1 + n= 1 [ n + 1 ) a n+ 1 2 a n 1 ] x n. 1 3) As a consequence, we have From this we clearly see that a 2 k + 1 = 0 for all k. On the other hand, Therefore we have Now we recognize that So finally we have a 1 = 0, a n+ 1 = 2 n + 1 a n ) a 2 k = 2 2 k a 2 k 2 = 1 k a 2 k 2 = = 1 a 0. k! 1 5) y x) = [ 1 a 0 x 2 k = a 0 k! k = 0 ] 1 x 2 k. k! 1 6) k = 0 1 x 2 k = 1 ) x 2 k = e x ) k! k! k = 0 k = 0 y x) = a 0 e x 2 1 8) where a 0 can take any value recall that the general solution to a first order linear equation involves an arbitrary constant! From this example we see that the method have the following steps: 1. Write y x) = a n x n. 1 9) 2. Substitute into the equation and determine a n. A recurrence relation a formula determining a n using a i, i < n is preferred. 3. Try to sum back and find out a closed form formula for y. There are several theoretical issues we need to settle. 1. When we substitute y x) = a n x n into the equation, we write y = a n x n ) 20) which is equivalent to claiming that it s OK to differentiate term by term: an x n ) = an x n ) 21 ) 2. We determine a n by settle the coefficients of each x n to 0. In other words, we claim that a n x n = 0 a n = 0 for each n. 22)

3 3. In practice, it may happen that we cannot sum back. Then there are two issues: a. Check whether a 0 + a 1 x + is indeed a well-defined function. b. If it is, but we cannot find a closed form formula, can we estimate how well the partial sum approximates the actual infinite sum? That is, any estimate of a n x n N a n x n? 23) In practice this is very important. For example, if we know that the difference decreases as 3/N 3 and the problem requires 2-digit accuracy, we know it suffices to sum up the first 1 0 terms. These issues are settled by the theory of power series and analytic functions Power series and analytic functions. A power series about a point x 0 is an expression of the form a n x x 0 ) n = a 0 + a 1 x x 0 ) + a 2 x x 0 ) ) Following our previous discussion, we want to know whether this infinite sum indeed represents a function, say F x), or not. If it does, then for any c we would have F c) = As the RHS is an infinite sum, it should be understood as This motivates the following. We say that exists and is finite. a n c x 0 ) n. 25) N F c) = lim N a n c x 0 ) n. 26) a n x x 0 ) n converges at the point x = c if the limit lim N N a n c x 0 ) n 27) If this limit does not exist, we say that the power series diverges at x = c. Clearly, equations like a n x n = 0 28) is only meaningful when the LHS converges. Recall that we would like to justify concluding a n = 0 from this equation. This is fulfilled by the following theorem. The most important property of power series is the following: Proposition 2. If the power series satisfying an x x 0 ) n is convergent at x = c, then it is convergent at all x x x 0 < c x 0. 29) From this property, it follows that If the power series diverges at c, then it has to diverge for all x such that x x 0 > c x 0. Combine these facts, we see that for any power series, there has to be a particular number ρ 0 such that the power series converges for all x such that x x 0 < ρ;

4 the power series diverges for all x such that x x 0 > ρ. This particular number ρ is called the radius of convergence of the power series. Remark 3. The number ρ is at least 0, as taking x = x 0 gives 0 which is clearly converging to 0; On the other hand, when the power series is convergent for all x, we say its radius of convergence is infinity, that is ρ =. Remark 4. Whether the power series converges at x = x 0 ± ρ is tricky to determine. Different approaches are needed for different power series. Inside x x 0 < ρ, life is perfect. In particular all our questions are satisfactorily answered. Theorem 5. Properties of power series inside x x 0 = ρ) Power series vanishing on an interval) If a n x x 0 ) n = 0 30) for all x in some open interval, then each a n is 0. 1 Differentiation and integration of power series) If the power series a n x x 0 ) n has a positive radius of convergence ρ, then if we set f x) to be the sum inside x x 0 < ρ, then f is differentiable and integrable inside this same interval. Furthermore we can perform differentiation and integration term by term. f x) = n a n x x 0 ) n 1 x x 0 < ρ; 31 ) n= 1 f x) dx = a n n + 1 x x 0) n+ 1 + C x x 0 < ρ. 32) Multiplication of power series) If a n x x 0 ) n has a positive radius of convergence ρ 1 and bn x x 0 ) n has a positive radius of convergence ρ 2, then termwise multiplication a 0 + a 1 x x 0 ) + ) b 0 + b 1 x x 0 ) + ) = a 0 b 0 + [ a 0 b 1 + a 1 b 0 ] x x 0 ) + 33) is OK for all x x 0 < ρ where ρ min { ρ 1, ρ 2 }. Remark 6. There are many alternative definition of the value of infinite sums which extends the perfect life outside x x 0 < ρ, that is they give meaning to a n x x 0 ) n for x x 0 = ρ or even x x 0 > ρ. Such techniques have practical values. In particular, in many cases the sum as defined by the alternative definitions) of a n x x 0 ) n for certain x with x x 0 > ρ still approximates the value of the solution. A few of such alternative summations are Cesaro summation, Abel summation, and Borel summation. Furthermore these new summations are closely related to the theory of asymptotic analysis. Example ) Find a power series expansion for f x), with Solution. We have f x) = 1 + x) 1 = 1 ) n x n. 34) f x) = [ 1 ) n x ] n = 1 ) n n x n 1 = 1 ) n+ 1 n + 1 ) x n. 35) n= 1 Note the change of index range in the 2nd equality. 1. The idea of radius of convergence is not explicit here. But notice that of the infinite sum, and consequently x x 0 ρ. an x x 0) n = 0 implies the convergence

5 Example ) Find a power series expansion for g x) = 0x f t) dt for Solution. Compute f x) = 1 + x) 1 = 1 ) n x n. 36) f x) = 1 ) n x n = 1 n ) x n+ 1 + C. 37) n + 1 The constant C is determined through setting x = 0: Therefore or if preferred, g x) = C = g 0) = 0. 38) 1 ) n n + 1 x n+ 1, 39) g x) = 1 n 1 ) x n. 40) n n= 1 Example ) Compute the power series 2 for f g with Solution. We first find the power series for f and g. f. We know that Thus f = e x, g = 1 + x) ) e x ) n) = 1 ) n e x. 42) e x = Note that the radius of convergence is. g. We have Therefore The radius of convergence is 1. 1 ) n n! ) 1 n) = 1 ) n n! 1 + x 1 + x) 1 = x n. 43) x) n. 44) 1 ) n x n. 45) Note that the radius of convergence for the power series of the product f g is the smaller of f and g. So in our case here is 1. Now we compute the product f g = = [ [ 1 ) n n! 1 x + x2 2 x3 6 + ] [ ] x n 1 ) n x n ] [ 1 x + x 2 x 3 + ] = 1 2 x x2 + 46) 2. When x 0 is not specified, it is implicitly set as 0.

6 Given that this number ρ is so important, we clearly would like to have a way to compute it. We have the following theorem. Theorem 1 0. Ratio test) If, for n large, the coefficients a n are nonzero and satisfy lim n a n+ 1 a n = L 0 L ) 47) then the radius of convergence of the power series a n x x 0 ) n is ρ = 1 /L, with ρ = if L = 0 and ρ = 0 if L =. That is, the power series converges 3 for x x 0 < ρ and diverges for x x 0 > ρ. Remark 1 1. In other words, if lim n a n+ 1 a n = L 0 L ) 48) then the power series coincide with a well-defined function for x x 0 < ρ = 1 /L. Example ) Determine the convergence set for Solution. We apply the ratio test. Compute Clearly we have a n+ 1 a n 2 n n + 1 x 1 ) n. 49) n n 1 n + 1 = 2 n+ 1 ) 1 lim n a n+ 1 a n = n n + 2). 50) = ) Thus the radius of convergence is 2 and the power series converges for 1 < x < 3 and diverges for x < 1 or x > 3. How about 1 and 3? For these points we need to determine in an ad hoc manner. x = 1. We have which is actually converging. 4 x = 3. We have which is diverging. 5 2 n n + 1 x 1 ) n = 2 n n + 1 x 1 ) n = 2 n n + 1 ) 2) n = 2 n n + 1 ) 2 n = 1 ) n n n + 1 Remark 1 3. As we will see soon, when solving ODEs using power series method, it is possible to know the radius of convergence before actually finding out the coefficients for the expansion of y. This is important knowing ρ enables us to estimate how well the partial sum approximates the solution, based on which we can determine how many coefficients we need to compute. 52) 53) 3. In fact, ab solutely converges. 4. an is converging if a n and a n + 1 have different sign for each n, lim a n = 0, and a n is decreasing n = 0 n x.

7 We have seen that the functions that can be represented by power series have very nice properties. Let s give them a name. Definition 1 4. Analytic functions) A function f is said to be x 0 if, in an open interval about x 0, the function is the sum of a power series a n x x 0 ) n that has positive radius of convergence. It turns out that this power series is exactly the Taylor expansion of f: with general term f x 0 ) + f x 0 ) x x 0 ) f x 0 ) x x 0 ) ) 1 n! f n) x 0 ) x x 0 ) n. 55) Remark 1 5. It would be very awkward if we have to determine analyticity of a function through the above definition. Fortunately we have other ways to do this. In particular, if f, g are analytic at x 0, so are c 1 f + c 2 g, f g, and f/ g when g x 0 ) 0). Remark 1 6. In the 1 8th century people believe all reasonable functions say continuous) are analytic. This clearly is the belief that fuels the research on power series method. Later in the 1 9th century it was discovered that this belief is wrong most functions do not equal to any good power series. Nevertheless, the method of power series survived this change and remains a powerful tool in solving ODEs even today More examples. Now we return to solving equations using power series. Example ) Find at least the first four nonzero terms in a power series expansion about x = 0 for a general solution to Solution. We write Substituting into the equation, we have z x 2 z = 0. 56) z x) = a 0 + a 1 x + = a n x n. 57) 0 = z x 2 z = a n n n 1 ) x n 2 n= 2 a n x n+ 2 = a n+ 2 n + 2) n + 1 ) x n n= 2 a n 2 x n = 2 a a 3 x + n= 2 [ a n+ 2 n + 2) n + 1 ) a n 2 ] x n. 58) Thus we have We conclude: 2 a 2 = 0 59) 6 a 3 = 0 60) a a n+ 2 n + 2) n + 1 ) = a n 2 a n+ 2 = n 2 n + 2) n + 1 ). 61 ) a 2 = 0 62) a 3 = 0 63) a 4 = a 0 64) 1 2 a 5 = a 1 65) 20

8 As we only need 4 nonzero terms, we stop here. The solution is z x) = a 0 + a 1 x + a x4 + a 1 20 x ) Example ) Find a power series expansion about x = 0 for a general solution to the given differential equation. Your answer should include a general formula for the coefficients. Solution. We write y x y + 4 y = 0. 67) y x) = a n x n. 68) Substituting into the equation, we have 0 = y x y + 4 y = a n n n 1 ) x n 2 x n= 2 n= 1 n a n x n a n x n = a n+ 2 n + 2) n + 1 ) x n n= 1 n a n x n + 4 a n x n = 2 a a 0 ) + n= 1 [ n + 2) n + 1 ) a n+ 2 n 4) a n ] x n. 69) This gives 2 a a 0 = 0 70) n + 2) n + 1 ) a n+ 2 n 4) a n = ) Therefore It is clear that we should discuss n = 2 k and n = 2 k 1 separately. For even n, we have For odd n, we have a 2 k + 1 = a 2 = 2 a 0, 72) a n+ 2 = n 4 n + 2) n + 1 ) a n. 73) a 2 = 2 a 0, a 4 = 1 6 a 2 = 1 3 a 0, a 6 = 0, a 8 = 0, a 1 0 = 0, 74) 2 k 5 2 k + 1 ) 2 k) a 2 k 5) 2 k 7) 2 k 1 = 2 k + 1 ) 2 k 2) a 2 k 3 = = 2 k 5) 3) a 1. 75) 2 k + 1 )! Summarizing, we have [ y x) = a x ] [ 3 x4 + a 1 x + 2 k 5) 3) 2 k + 1 )! k = 1 x 2 k + 1 ]. 76) Example ) Find at least the first four nonzero terms in a power series expansion about x = 0 for the solution to the given initial value problem. Solution. We write w + 3 x w w = 0, w 0) = 2, w 0) = 0. 77) w x) = a n x n. 78)

9 Substituting into the equation, we obtain 0 = n= 2 a n n n 1 ) x n x n= 1 n a n x n 1 a n x n = a n+ 2 n + 2) n + 1 ) x n + n= 1 3 n a n x n a n x n = 2 a 2 a 0 + n= 1 [ n + 2) n + 1 ) a n n a n a n ] x n. 79) Therefore which leads to 2 a 2 a 0 = 0 80) n + 2) n + 1 ) a n n 1 ) a n = 0 81 ) a 2 = 1 2 a 0 82) a n+ 2 = On the other hand, the initial values give Therefore we can compute successively 1 3 n n + 2) n + 1 ) a n. 83) 2 = w 0) = a 0, 0 = w 0) = a 1. 84) a 2 = 1 2 a 0 = 1, 85) a 3 = 2 a 1 = 0, 6 86) a 4 = a 2 = 5 1 2, 87) a 5 = 8 20 a 3 = 0, 88) a 6 = a 4 = ) We stop here as only four nonzero terms are required. Finally the answer is w x) = 2 + x x x ) Ordinary and singular points. Not all linear ODEs are amenable to the above naïve approach: Write expansion Substitute into equation Determine coefficients. Example ) Consider Solution. Write Substituting into the equation, we obtain 0 = x 2 n= 2 x 2 y + 3 y x y = ) y = a n n n 1 ) x n n= 1 a n x n. 92) n a n x n 1 a n x n+ 1 = n= 2 a n n n 1 ) x n + 3 n + 1 ) a n+ 1 x n n= 1 a n 1 x n = 3 a a 2 a 0 ) x + n= 2 [ a n n n 1 ) + 3 n + 1 ) a n+ 1 a n 1 ] x n. 93)

10 This leads to which leads to 3 a 1 = 0 94) 6 a 2 a 0 = 0 95) 3 n + 1 ) a n+ 1 + n n 1 ) a n a n 1 = 0, n 2 96) a 1 = 0, a 2 = a 0 6, a n+ 1 = a n 1 n n 1 ) a n. 97) 3 n + 1 ) Clearly, all coefficients are determined once a 0 is given. In other words, the power series solution we obtain has only one arbitrary constant. As the equation is of second order, this means not all solutions can be obtained through expansion into power series. Checking the difference between this example and the previous ones, we reach the following definition. Definition 21. Ordinary and singular points) Consider the linear differential equation in the standard form y + p x) y + q x) y = 0. 98) A point x 0 is called an if both p, q are analytic at x 0. If x 0 is not an ordinary point, it is called a of the equation. Example 22. Find the singular points of the equation we just discussed. Solution. First write it into the standard form x 2 y + 3 y x y = 0. 99) We have y + 3 x 2 y 1 y = ) x p x) = 3 x 2, q x) = 1 x ) As 3, 1, x 2 and x are all analytic everywhere, their ratios are analytic at all points except those making the denominator vanish. Therefore the only singular point is x = 0. Remark 23. When x 0 is an ordinary point, we expect no trouble finding solutions of the form y x) = a n x x 0 ) n. 1 02) When x 0 is singular, in general not all solutions can be represented in the above form, as we have seen in the example. 2. Equations with analytic coefficients. The following theorem summarizes the good properties of ordinary points. Theorem 24. Existence of analytic solutions) Suppose x 0 is an ordinary point for equation y x) + p x) y x) + q x) y x) = 0, 1 03) then it has two linearly independent analytic solutions of the form y x) = a n x x 0 ) n. 1 04) Moreover, the radius of convergence of any power series solution ofthe form given above is at least as large as the distance from x 0 to the nearest singular point real or complex-valued).

11 Remark 25. points. It is important, when determine the radius of convergence, to count in complex singular Remark 26. Recall that, for many problems, we could not write down an explicit formula for the coefficients a n, but can only compute a few terms. For example, when considering we obtain w + 3 x w w = 0, w 0) = 2, w 0) = ) w x) = 2 + x x x ) With the help of this theorem, we can estiamte how good the first few terms approximates w x), that is we can estimate the size of w x) 2 + x x ) 72 x ) Example 27. about x ) Find a minimum value for the radius of convergence of a power series solution x + 1 ) y 3 x y + 2 y = 0, x 0 = ) Solution. Write the equation to standard form y 3 x x + 1 y + 2 x + 1 y = ) The only singular point is x = 1. Thus the minimum radius of convergence is the distance between x 0 = 1 and 1, which is 2. Example 28. about x ) Find a minimum value for the radius of convergence of a power series solution 1 + x + x 2 ) y 3 y = 0; x 0 = ) Solution. Write the equation to standard form y x + x 2 y = ) The singular points are roots of 1 + x + x 2, which are x 1, 2 = 1 ± 3 i ) 2 To find out the closest singular point to x 0, we compute x 1 x 0 = i = 3 ; x2 x 0 = ) So both are 3 away from x0. As a consequence, the minimum radius of convergence is 3. Example ) Find at least the first four nonzero terms in a power series expansion about x 0 for a general solution to the given differential equation with the given value of x 0. x 2 2 x ) y + 2 y = 0, x 0 = ) Solution. The best way to do this is to first shift x 0 to 0. To do this, let t = x 1. Then t 0 = x 0 1 = 0, x 2 2 x = t 2 1, and the equation becomes t 2 1 ) y + 2 y = ) and we would like to expand at t 0 = 0.

12 Substituting y = a n t n 1 1 6) into the equation, we have 0 = t 2 1 ) a n n n 1 ) t n a n t n n= 2 = n= 2 a n n n 1 ) t n n= 2 a n n n 1 ) t n a n t n = n= 2 a n n n 1 ) t n a n+ 2 n + 2) n + 1 ) t n + 2 a n t n = 2 a a 0 ) + 6 a a 1 ) t + n= 2 [ a n n n 1 ) a n+ 2 n + 2) n + 1 ) + 2 a n ] t n ) This leads to 2 a a 0 = ) 6 a a 1 = ) a n n n 1 ) a n+ 2 n + 2) n + 1 ) + 2 a n = ) As only four terms are needed, all we need to settle is Thus a 2 = a 0, a 3 = 1 3 a ) y t) = a 0 + a 1 t + a 0 t a 1 t ) Back to the x variable, we have y x) = a 0 [ 1 + x 1 ) 2 + ] + a 1 [ x 1 ) x 1 ) 3 + ]. 1 23) Example ) Find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem. y sin x) y = 0, y π) = 1, y π) = ) Solution. As the initial values are given at π, the expansion should be about x 0 = π. First introduce new variable t = x π. Then x = t + π and the equation becomes Now write and substitute into the equation, recalling we reach y + sin t) y = 0, y 0) = 1, y 0) = ) sin t = y t) = a n t n 1 26) 1 ) n t 2 n+ 1, 1 27) 2 n + 1 )! 0 = y + sin t) y = a n n n 1 ) t n 2 + t 1 ) ) t 3 + a0 + a 1 t + a 2 t 2 + a 3 t 3 + 3! n= 2 = 2 a a 3 t a 4 t a 5 t a 0 t + a 1 t 2 + a 2 a ) 0 t = 2 a a 3 + a 0 ) t a 4 + a 1 ) t a 5 + a 2 a ) 0 t ) 6

13 This gives Applying the initial values we have Thus the above leads to 2 a 2 = ) 6 a 3 + a 0 = ) 1 2 a 4 + a 1 = ) 20 a 5 + a 2 a 0 6 = ) a 0 = 1, a 1 = ) a 0 = 1, a 1 = 0, a 2 = 0, a 3 = 1 6, a 4 = 0, a 5 = ) Only three of them are non-zero. We have to return to the equation and expand to higher orders. 0 = y + sin t) y = a n n n 1 ) t n 2 + t 1 3! n= 2 t ! t 5 + ) a0 + a 1 t + a 2 t 2 + a 3 t 3 + a 4 t 4 ) which gives = 2 a a 3 t a 4 t a 5 t a 6 t 4 + a 0 t + a 1 t 2 + a 2 a ) 0 t 3 + a 3 a ) 1 t ) 6 6 from balancing the t 4 term. As a 1 = 0, a 3 = 1 /6, we have 0, we already have 4 nonzero terms and do not need to go to higher orders. The solution in t vari- As a 6 able is 30 a 6 + a 3 a 1 6 = ) a 6 = ) Returning to the x-variable, we have y t) = t t t ) y x) = x π) x π) x π) ) There is no difficulty extending the power series method to non-homogeneous problems. Example ) Find at least the first four nonzero terms in a power series expansion about x = 0 of a general solution to the given differential equation. Solution. Write y x y + 2 y = cos x. 1 40) y x) = a n x n 1 41 ) and substitute into the equation, recalling cos x = 1 ) n 2 n)! x 2 n, 1 42)

14 we have 1 ) n 2 n)! x 2 n = cos x = y x y + 2 y 2 a n x n = n= 2 = a n n n 1 ) x n 2 n= 1 a n+ 2 n + 2) n + 1 ) x n n= 1 n a n x n + n a n x n + 2 a n x n = 2 a a 0 ) + n= 1 [ a n+ 2 n + 2) n + 1 ) n a n + 2 a n ] x n. 1 43) The first few terms of balance is then: From these we have So the solution is x 0 : 1 = 2 a a ) x 1 : 0 = 6 a 3 + a ) x 2 : 1 2 = 1 2 a ) a 2 = 1 2 a 0, a 3 = 1 6 a ) ) 1 y x) = a 0 + a 1 x + 2 a 0 x a 1 x 3 + ) 1 ) = 2 x2 + + a 0 1 x a1 Remark 32. From the above we can conlude that the expansions of y p, y 1, y 2 are Don t forget that the choices of y p, y 1, y 2 are not unique! ) x 1 6 x3 + ). 1 48) y p x) = 1 2 x ) y 1 x) = 1 x ) y 2 x) = x 1 6 x ) 3. C auchy-euler equations. In the above we have seen that, using power series about an ordinary point, we can easily obtain the power series representations for the general solution. On the other hand, if we expand about a singular point, not all solutions can be obtained through power series. A typical equation with singular point is the Cauchy-Euler equation a x 2 y x) + b x y x) + c y x) = 0, x > ) where a, b, c are constants. We have studied this equation before, the conclusions are To find the solutions, we need to consider the associated characteristic, or indicial, equation which can be obtained by substituting y = x r into the equation. There are three cases. a r 2 + b a) r + c = ) 1. The characteristic equation has two distinct real roots r 1, r 2. Then the general solution is given by 2. The characteristic equation has one double root r = r 0. Then y = c 1 x r1 + c 2 x r ) y = c 1 x r0 + c 2 x r0 ln x. 1 55)

15 3. The characteristic equation has two complex roots α ± i β. Then y = c 1 x α cos β ln x) + c 2 x α sin β ln x). 1 56) From these we clearly see that, unless r 1, r 2 are both non-negative integers, there is no way that we can solve the equation by setting y x) = a n x n, 1 57) since none of the solutions would be analytic around 0. We also observe that in case 2, the second linearly independent solution, x r0 ln x, can be obtained formally from the first one by differentiating with respect to r: x r0 ln x = ln x) e r0 ln x = [ ] e r ln x r = r r 0 = r xr ) r = r ) To understand why this would work, recall that a x r ) + b x r ) + c x r = [ a r 2 + b a) r + c ] x r 1 59) When the characteristic equation has a double root r 0, the above becomes Now differentiating both sides with respect to r, we have a x r ) + b x r ) + c x r = a r r 0 ) 2 x r. 1 60) a x r ln x) + b x r ln x) + c x r ln x) = 2 a r r 0 ) x r + a r r 0 ) 2 x r ln x ) Setting r = r 0, we see that x r0 ln x is indeed a solution. The above observations play important roles in the so-called Method of Frobenius for solving problems at singular points using power series. This method is discussed in 8. 6 and Example ) Find general solution. x 2 y x) + 6 x y x) + 6 y x) = ) Solution. We try y = x r. Substituting into the equation, we reach the characteristic equation Thus the general solution is r r + 6 = 0 r 1, 2 = 2, ) y x) = c 1 x 2 + c 2 x ) Example ) Find the general solution for x 2 y x) x y x) y x) = ) Solution. Try y = x r. We reach the characteristic equation r 2 2 r = 0 r 1, 2 = 1 ± 4 i 1 66) Thus the general solution is y x) = c 1 x cos 4 ln x) + c 2 x sin 4 ln x). 1 67) There is no problem generalizing to higher order equations. Example ) Find the general solution for x 3 y x) + 4 x 2 y x) x y x) 1 0 y x) = ) Solution. Try y = x r. The equation becomes [ r r 1 ) r 2) + 4 r r 1 ) r 1 0] x r = )

16 which gives the characteristic equation r r 1 ) r 2) + 4 r r 1 ) r 1 0 = ) Note that we didn t expand everything. The reason is, we notice that 1 0 r 1 0 = 1 0 r 1 ) so the LHS can be conveniently factorized: LHS = r 1 ) [ r r 2) + 4 r + 1 0] = r 1 ) r r ) ) The roots for r r are 1 ± 3 i. Thus the three roots are The corrsponding linearly independent solutions are The general solution is then given by r 1 = 1, r 2, 3 = 1 ± 3 i. 1 72) x, x 1 cos 3 ln x), x 1 sin 3 ln x). 1 73) y x) = c 1 x + c 2 x 1 cos 3 ln x) + c 3 x 1 sin 3 ln x). 1 74) Recall the derivation of the method of variation of parameters: Given Let y 1, y 2 be solutions to the homogeneous equation We try to find a particular solution of the form Substituting into the equation, we have a y + b y + c y = g. 1 75) a y + b y + c y = ) y = v 1 y 1 + v 2 y ) a [ v 1 y v 1 y 1 + v 1 y 1 + v 2 y v 2 y 2 + v 2 y 2 ] + b [ v 1 y 1 + v 1 y 1 + v 2 y 2 + v 2 y 2 ] + c [ v 1 y 1 + v 2 y 2 ] = g. 1 78) Now a y 1 + b y 1 + c y 1 = 0 a v 1 y 1 + b v 1 y 1 + c v 1 y 1 = ) and a similar relation holds for v 2, y 2. Using these many terms can be cancelled and we reach a [ v 1 y v 1 y 1 + v 2 y v 2 y 2 ] + b [ v 1 y 1 + v 2 y 2 ] = g. 1 80) Now we require This leads to v 1 y 1 + v 2 y 2 = ) v1 y 1 + v 2 y 2 ) = v1 y 1 + v 1 y 1 + v 2 y 2 + v 2 y 2 = ) and cancels another 6 terms. We thus obtain the following system for v 1, v 2 : y 1 v 1 + y 2 v 2 = g/a 1 83) y 1 v 1 + y 2 v 2 = ) Solving it we obtain v 1 = g y 2 a y 1 y 2 y 1 y 2 ), v 2 g y = 1 a y 1 y 2 y 1 y 2 ). 1 85) Now checking the above process, we see that nowhere have we used the fact that a, b, c are constants. In other words, the formulas for v 1, v 2 still hold for variable coefficient equations!

17 Example ) Use variation of parameters to find a general solution to the given equation for x > 0. x 2 y x) 2 x y x) + 2 y x) = x 1 / ) Solution. We first solve the homogeneous equation, then use variation of parameters. 1. Solve the homogeneous equation The characteristic equation is So 2. Variation of parameters. First compute Thus we have 3. Obtain general solution. The general solution is then x 2 y x) 2 x y x) + 2 y x) = ) r 2 3 r + 2 = 0 r 1 = 2, r 2 = ) y 1 = x 2, y 2 = x. 1 89) a y 1 y 2 y 1 y 2 ) = x 2 x 2 2 x 2 ) = x ) v 1 = x1 / 2 x 4 = x 7/ 2 v 1 = 2 5 x 5 / ) v 2 = x3/ 2 x 4 = x 5 / 2 v 2 = 2 3 x 3/ ) y = c 1 y 1 + c 2 y 2 + v 1 y 1 + v 2 y 2 = c 1 y 1 + c 2 y x 1 / x 1 / 2 = c 1 y 1 + c 2 y x 1 / )