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490 DIFFERENTIAL EQUATIONS [ ] y = Ae x + Be −2x − 4 3 Now try the following exercise. − 2 1 x − 2 1 x2 + 4 1 e2x Exercise 193 Further problems on second d 2 y order differential equations of the form 4. a d2 y dx 2 + b dy dt 2 − 2dy dt + 2y = et sin t [ + cy = f (x) where f (x) is a sum y = e t (A cos t + B sin t) − 2 t dx et cos t ] or product In Problems 5 to 6 find the particular solutions of the given differential equations. In Problems 1 to 4, find the general solutions of the given differential equations. d 2 y 5. dx 1. 8 d2 y 2 − 7dy dx + 10y = e2x + 20; when x = 0, dx 2 − 6dy + y = 2x + 40 sin x y = 0 and dy dx dx =−1 3 ⎡ ⎤ [ ⎣ y = Ae 4 x + Be 2 x + 2x + 12 + 8 y = 4 17 (6 cos x − 7 sin x) ⎦ 3 e5x − 10 3 e2x − 1 ] 3 xe2x + 2 d 2 6. 2 d2 y y 2. dθ 2 − 3dy dx 2 − dy dx − 6y = 6ex cos x; when x = 0, + 2y = 2 sin 2 θ − 4 cos 2 θ dθ y =− 21 dy and [ 29 dx =−620 29 y = Ae 2θ + Be θ + 2 1 ( sin 2 θ + cos 2 θ)] ⎡ ⎤ d 2 y 3. dx 2 + dy ⎣ y = 2e− 2 3 x − 2e 2x dx − 2y = x2 + e 2x + 3ex ⎦ 29 (3 sin x − 7 cos x)
Differential equations 52 Power series methods of solving ordinary differential equations 52.1 Introduction Second order ordinary differential equations that cannot be solved by analytical methods (as shown in Chapters 50 and 51), i.e. those involving variable coefficients, can often be solved in the form of an infinite series of powers of the variable. This chapter looks at some of the methods that make this possible—by the Leibniz–Maclaurin and Frobinius methods, involving Bessel’s and Legendre’s equations, Bessel and gamma functions and Legendre’s polynomials. Before introducing Leibniz’s theorem, some trends with higher differential coefficients are considered. To better understand this chapter it is necessary to be able to: (i) differentiate standard functions (as explained in Chapters 27 and 32), (ii) appreciate the binomial theorem (as explained in Chapters 7), and (iii) use Maclaurins theorem (as explained in Chapter 8). 52.2 Higher order differential coefficients as series The following is an extension of successive differentiation (see page 296), but looking for trends, or series, as the differential coefficient of common functions rises. (i) If y = e ax , then dy dx = aeax , d2 y dx 2 = a2 e ax , and so on. If we abbreviate dy dx as d 2 y y′ , dx 2 as y′′ , … and d n y dx n as y(n) , then y ′ = ae ax , y ′′ = a 2 e ax , and the emerging pattern gives: y (n) = a n e ax (1) For example, if y = 3e 2x , then d 7 y dx 7 = y(7) = 3(2 7 )e 2x = 384e 2x (ii) If y = sin ax, ( y ′ = a cos ax = a sin ax + π ) 2 y ′′ =−a 2 sin ax = a 2 sin (ax + π) ( = a 2 sin ax + 2π ) 2 y ′′′ =−a 3 cos x ( = a 3 sin ax + 3π 2 In general, For example, if ) and so on. ( y (n) = a n sin ax + nπ 2 y = sin 3x, then d5 y dx 5 = y(5) ( = 3 5 sin 3x + 5π ) ( = 3 5 sin 3x + π ) 2 2 = 243 cos 3x ) (2) (iii) If y = cos ax, ( y ′ =−a sin ax = a cos ax + π ) 2 ( y ′′ =−a 2 cos ax = a 2 cos ax + 2π ) 2 ( y ′′′ = a 3 sin ax = a 3 cos ax + 3π ) 2 and so on. ( In general, y (n) = a n cos ax + nπ ) (3) 2 I
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Page 15 and 16: Hence from equation (4): ye x = xe
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