Question

PLEASE JUST DO QUESTION 1

Differential Equations // Math 2680 Section OL63 // Summer 2020 Using Power Series For each differential equation P(x)/' + Q(x) +R()y= 0, the solution will be a power series y(x) = n(1-10)" = 10 +0,1 + aza? +... where 20 satisfies P(x0) +0, and ao = A and a - B are constants that determine all other coefficients. 10 0 NO Example: " + y = 0. Let Xo = 0. Plug in y(x) = " into the differential equation y' + y = (4,2")" Canz") - ( Ën(n − 1)ancn-2) + (Ŝancz") ((n + 2)(n + 1)an +22") + (Ženz") – Ž ((n +2)(n + 1)an +2 + an )z" = 0 (n + 2)(n + 1)an+2 + anz". Rearranging we get On+2= (n+2)(+1), which says 22 =20,04 als , and a6 = With = 70. Also, a3 = =, as = d = 1 and ay = m = 56. Therefore y(I) = 40 + aj1+(99)? ++)+(93) + (%)25 +(33)2 + (506)z? +... 22 23 27 ...) + az (= 5x +...) Therefore 0 6 + .. + 6 120 5040 For the following differential equations, find 42, 43, 44, 45, 46, and ay in terms of do and a, and write the answer y(x) = 20 sum of terms sum of terms :) + +01 1. 1-0 expanding about #0 -0.

Answer 1

o aiven differential canation (DE) is y"_MEO The power sexies solution to about Xoro y q MED anan Then y = E nennilan an-2 and nanana no n=2 substitute y, y and all in . Then y"-yao - nem is an an-2 Ž anana nel no E (2+2) Entisanta sen § anah so no neo Ź ((n+1}{n+eJan+2 - an] =* =0 NO which is an identity, equating the coeffklents of various Powers of x" to zero, we get (hti (n+2) anta - an o nzo an no ante Inti) (hA2) which is known os Recurrence Relation for no for nal; Qq = £ ao = 옳 Qy=ta = 1/2 az = fa rao ->945 / 20 For n=2; ац

For n=3; as = 1/2 az sa, По я Есҡ те + , aь an 1. wo о > aь = Go and Sp Ә., substitute aoi ajari О) а 1 QS, Qы. values in @ 2 = q+q, x+a, x-qa> оцм*4+q25+ > = a +9х+++59,7°4 oz++ 54, 5, оф. >= (1+ + + — ) +(x+ -Ja, + + क qol 1+ ая — ) +a tuz +zab 6! 4 4. Required solution