MT171: Question Sheet 9-2018/19: Hints to solutions 1 (a) Integrating by parts with u = (a2 - t2)" and du = 1, In =(a2-t2) n dt = [t(a2-t2) ]0 - | t(-2nt(a2 -+2)n-1 dt 0 = 2n +2 (a2 - +2)n- 1 dt 0 ra Therefore (2n + 1)In = 2na2In-1 . ra = - 2n |(a2-+2)n dt - | a2 (a2 -12)n-1 dt]. 6a2 Hence I3 = 12 = 6a2 4a2 7 5 I1 = 6a2 4a2 2a2 7 5 3 1. 1 dt = 16a7 35 . (b) Integrating by parts with u = (1 - x)" and du = coshx, In = (1 - x)" cosh a dx = [(1 - x)" sinh x], + / sinhx[n(1-x)"-1] dx Jo = n | sinhx(1-x)"-1 dx. Integrating by parts again with u = (1 - x)"-1 and du = sinh x and using n ? 2, 1 0 ·1 In =|(1 -x)" cosh x dx = n [(1 - x)~1 cosh x]| - |cosh .[-(n - 1)(1 - x)2-2] dx =- n+n(n-1)In-2 , i.e. In +n = n(n-1)In-2 . Hence I4= - 4+ 1212 whilst = =- 4+12[-2+2Io] -28 +24/ cosh x dx = 24 sinh 1 - 28, I5 = - 5+ 20I3 =- 5+20[-3+6/1] ~1 = 120 / (1- x) cosh x dx - 65 0 = 120([(1 - x) sinh x]' + | sinh a dx) - 65 = 120[cosh 1-1] - 65 = 120 cosh 1 - 185.
2. Separating the variables: (a) 1 1 + y2 dx dy 1+ x2 1 i.e. arctan y = arctan x + arctanc, say giving y = tan(arctan x + arctan c) = x+C 1 - cx ' (b) for arbitrary constant c, using the given hint. y cos y dx dy 1+ x2 2x2 i.e. ysiny - | sin v dv = (2 - 1 + u2 2 du i.e. ysiny + cos y = 2x - 2 arctan x + c , for arbitrary constant c. (c) dy dx =(1-x)(1 + y2) 1 1+ y2 dx dy =1-x so that giving arctany = x - " + c , 1 for arbitrary constant c. Using the boundary condition, arctan 1 = 1 - 5 + c 3.(a) giving == = + c arctany = x - +- dy + xy = x3 dx giving the integrating factor p(x) = exp| tdt] = exp[2], 2 to within a multiplicative constant. Then dx ex2 /2 dy + xe+212 y = x3 e=2 /2 i.e. T lye 2/2] = 23 0 2/2 integrating w.r.t. x yex2/2 = |+2 x te12/2 i.e. yet2 12 = [+2 et2/2]2 - 2 | tet2/2 dt = (x2 - 2)eª2/2 + c or y= x2 - 2+ ce-x2/2 ,
+ (cot x)y = 2cosecx for arbitrary constant c. (b) dy dx giving the integrating factor u(x) = exp[/ cottdt] = exp[Insin x] = sinx, to within a multiplicative constant. Then sin x dy + y cos x = 2 dx i.e. d dx -[y