Question

ASK YOUR TEACHER Sy, find an expression for Consider the following differential equation to be solved using a power series. y" - y = 0 Using the substitution y = in terms of ck+ 1 for k = 0, 1, 2, .... no C+2 k+ 2 = + + + + 4 + + V2 = x - . + Y2 + + Find two power series solutions of the given differential equation about the ordinary point x = 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solution. O Y = 1 + and 2! x x 4! Y2 = x+ 6! 3! 5! 7! x2 O yn = 1 - and x? 2: 4! 6! 3! 5! 7! O y = x and Y2 = = 1 + x + 21 3 O yn = 1 + x and x² + x3 + x 2! 31 4 5! O Y = 1 and x2 + x3 21 31 4! These solutions are recognized as O v1 = cos(x) and Y2 = sinh(x) Oyn = 1 + x and Y2 - -1 - x + et Ovo = * and Y2 = 0 Oyn = 1 and y2 = -1 + et O Y = cos(x) and Y = sin(x) Y₂=x+ + The general solution obtained from these two solutions is y(x) = obtained using the method of Section 4.3. which is equivalent to the solution

Answer 1

0 Given differential Equation is y" - y'=0 let us use the substitution. y = { Calm * nco This implies y'= 2 nnen- n=1 and مه 3 n(n-1) an xn-2 3 no 2 Using ® & in 0, we Get 8 n(n-D (n & M-2 n Corona n=2 = 0 n=1 U Take n-2=k Take n-1=k Then (x+2) (K+1) CK+2 xk (k+1) Ckti nt = 0. k=o K=o or [(+2)(k+d) x+2 (k+1) CK+1 -70* = k=o or (k+2)(x+1) C#+2 – (<+1) Ck+1 = (x+2)(x+1) Cx+2 (k+1) kez G+1 or - CK+2 = Сk+ ,K= 0,1,2,

2 Then, IC, - (for k= o) 2 j 2! L G C C, (for k=1) C3 = ģes CL959 3 3! G (for kez) and so on. N Now, from ya Cnan neo - to tanto x² + x + aut. 11 Co tant | qu? + ㅗ 3! G 23 + i Gu" + -Cota (* at x² + x3 teu to 2! 3! 41 or y = . (1) + c ( ។ nt a² 2! tu3 ta 3! ui + On Comparing it with y = coy,+ q Ya, we have Y,= 1 and 42 at n 21 2 + toms + 244 +. ui 3! ** or Y = 1 and Y2 = -1 + I +r+ x2 + 2! 3! " + or Y, = 1 and Y2 = -ten

3 Thus, Ok * can also be recognised as ہو Y = 1 Y2=-1ten