Solve y\'\'-y=0 By power series expansion Show the answer and show the series ex
ID: 1892314 • Letter: S
Question
Solve y''-y=0 By power series expansionShow the answer and show the series expansion for Sinh(x) and Cosh(x) based on your findings
Explanation / Answer
write y = Sum[n=0 to inf, a(n) x^n ], where the constants a(n) are to be determined. supposing we can differentiate term by term, we have: y ' = Sum[n=1 to inf, n a(n) x^(n-1) ] = Sum[n=0 to inf, (n+1) a(n+1) x^n ] = a(1) + Sum[n=1 to inf, (n+1) a(n+1) x^n ] but we also have: - 2 xy = - 2x Sum[n=0 to inf, a(n) x^n ] = Sum[n=0 to inf, - 2a(n) x^(n+1) ] = Sum[n=1 to inf, - 2a(n-1) x^n ] and so, equating term by term: a(0) is yet unknown; a(1)=0 and in general: (n+1) a(n+1) = - 2a(n-1) for n>=1 or n a(n) = - 2 a(n-2), for n>1 since a(1)=0, it's easy to see that a(2n+1)=0, ie all the odd terms are zero. for the even terms, 2n a(2n) = - 2 a(2n-2), for n>0, or: n a(2n) = - a(2n-2) for n>0. So we have a(2) = - a(0), a(4) = 1/2 a(0), a(6) = - 1/(3*2) a(0), a(8) = 1/(4*3*2) a(0), and we could verify by induction that we have: a(2n) = (-1)^n a(0) / n! (I'll leave this to you). So we can write back: y = Sum[n=0 to inf, (-1)^n a(0) x^(2n) / n! ] = a(0) Sum[n=0 to inf, ( - x^2 )^n / n! ] we recognize the exponential series for -x^2: y = a(0) exp( - x^2) ie: y = a(0) e^( - x^2) Actually, the differential equation is easier directly: y ' = - 2 x y, ie: dy / y = - 2 x dx so: ln y = - x^2 + constant, or y = k exp ( - x^2), k a constant.