Compute the first 6 coefficients aoa5 Please show all work S

Compute the first 6 coefficients (a_o-a₅)

Please show all work

Solve by using power series: 2y\'-y=sin h(x) Compute the first 6 coefficients (a_o-a_5)

Solution

We assume that there exists a solution to the D. E. which can be represented by a Taylor series. This is not always the case. We write y = a0 + a1x + a2x2 + ··· + akxk + ak+1xk+1 + ak+2xk+2 + ... Then y = a0 + a1x + ··· + akxk + ak+1xk+1 + ··· y0 = a1 + 2a2x + ··· + kakxk1 + (k + 1)ak+1xk +(k + 2)ak+2xk+1 + ··· y00 = 2a2 + 3 · 2a3x + ... +(k + 2)(k + 1)ak+2xk + ··· 2xy0 = 2a1x ··· 2kakxk ··· 2y = 2a0 2 a1x + ··· 2 akxk ··· Substituting in the equation y00 2xy0 2y = 0 yields 2a2 2a0 = 0, 6a3 4a1 = 0, (k + 2)(k + 1)ak+2 2(k + 1)ak = 0. Thus ak+2 = 2ak/(k + 2). Substituting r = k 2 yields ar = 2ar2/r . Therefore, for instance a8 = 2a6 8 = 22a4 8 · 6 = 23a2 8 · 6 · 4 = a2 4 · 3 · 2 = a0 4 · 3 · 2 , and in general, for any s 2, a2s = 2a2s2 2s = 4a2s4 4s(s 1) = 8a2s6 8s(s 1)(s 2) = ··· = 2sa0 2ss! = a0 s! and for s 1, a2s+1 = 2a2s1 2s + 1 = 22a2s3 (2s + 1)(2s 1) = ··· = 2sa1 (2s + 1)(2s 1)··· 3 . Thus y = a0 X s=0 x2s s! + a1 X s=0 2sx2s+1 (2s + 1)··· 3 · 1 = a0ex2 + a1 X s=0 2sx2s+1 (2s + 1)··· 3 · 1 where a0 and a1 are arbitrary constants.