Please make sure to answer the entire question, or else your answer will be disliked and reported for spam. If you belie

[answerhappygod]

Please make sure to answer the entire question, or else your
answer will be disliked and reported for spam. If you believe that
the question is too much to handle, don't answer it all. If you
think you can't read the questions properly because it's unclear,
don't answer it or leave a comment, and just move on. I've had
problems like these in the past and I will report you for spam in
this case too. Once you've fully answer the question, make sure
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The only reason why I'm making these disclaimers is because I've
had these problems in the past. I don't have the intention of being
nagging on unimportant details. As long as you follow these basic
requests, you'll get your like.

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Answers for Part A to Part C

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Consider the second-order differential equation y" – 4y= 0, y(0) = ao , y(0) =a1 a) Using the power series ansatz y (x) = P on=an Xn show the recurrence relation an+2 = 4an / (n+ 2)(n+ 1), n = 0, 1, 2, ... b) Determine power series solutions y1, y2 near xo= 0 such that y (x) =aoy1(x) +a1y2(x) using the recurrence relation (no verification necessary). c) Show that y1(x) = cosh(2x) =(e^(2x)+e^(-2x))/ 2, y2(x) =1/2 sinh(2x) =(e^2x-e-A(2x))4 using the Taylor expansions cosh(z) = .£Xn=0 z2n/ (2n)!, sinh(z) = .EXn=0 (Z2n+1)/(2n+ 1)! What is the radius of convergence of the series solutions y1, y2? (Hint: You may use the fact that the functions sinh(z), cosh(z) have no singularities in the complex plane

Part A) | 4an ((η + 2) (η + 1)αη+2 – 4αη = 0 και αη+2 = (η + 2)(η +1) Part B) y(α) = αο - 4kg2k 4kg2k+1 (2k)! +αι Σ (2k + 1)! + k=0 k=0 yı(x) Y2(c)

Part C) 91(a) = 44,22 cosh(2x) 22k x2k (2x)2k (2k)! (2k)! (2k)! k=0 k=0 k=0 4k x2k+1 1 22k+1x2k (2k + 1)! 2 (2k + 1)! k=0 1 y2(x) IM: (2x)2k+1 (2k + 1)! sinh(22). 2 k=0 k=0