Recent content by Mark Brewer

Okay. Thank you, I'll reply as soon as I see my mistakes.

A polynomial would then be formed, right?

sorry, I'm not sure if I can split the roots to two rationals.

I'm not sure if splitting the roots, or can I?

Homework Statement L-1{(2s2+3)/(s2+3s-4)2} The Attempt at a Solution I factored the denominator f(t)=(2s2+3)/((s-1)(s+4))2 now I've tried partial fractions to get (2s2+3)/((s-1)(s+4))2 = A/(s-1)2 + B(s+4)2 (2s2+3)=A(s+4)2 + B(s-1)2 by substitution, s=1 and s=-4 5=A(25) A=1/5 35=B(25)...

Thank you, Greg. You can bump this post.

Homework Statement Show that the Bessel functions Jn(x) (where n is an integer) have a very nice generating function, namely, G(x,t) := ∑ from -∞ to ∞ of tn Jn(x) = exp {(x/2)((t-T1/t))}, Hint. Starting from the recurrence relation Jn+1(x) + Jn-1(x) = (2n/x)Jn(x), show that G(x,t)...

I'm not sure why the second half of my problem is in a subscript. The double checked the problem by previewing it, but I could get the text out of being in a subscript. My apologies for this.

Homework Statement Starting from the recurrence relation, show that, when l is an integer, the polynomial solution to Legendre's equation is yl(x) = Kl ∑ from k = 0 to (l/2) of (((-1)k) / k!) (((2l - 2k)!) / (l-k)! (l - 2k)!) (xl-2k) where Kl is an arbitrary constant (depending on l) and x...

My new answer is extremely long. y = c1sinx + c2cosx + sin2x - xcosx(xsinx - 1) - lntanx + secx - cosx(sinxtanx - 2) -xsinx(cosx + 1)

Hi Ehild, My apologies for thinking I could use sec2x - 1. I thought the derivation was tanx from sec2x. I'll use sinx/cosx. Thank you again for the help.

Thank you, the secant should have been sec2x -1

Homework Statement (Reduction of order) The function y1 = x-1/2cosx is one solution to the differential equation x2y" + xy' + (x2 - 1/4) = 0. Use the method of reduction of order to find another linearly independent solution. The Attempt at a Solution I divided x2 to both sides to get the...

W = 1 r = tan2x y1 = sinx y2 = cosx yh = c1sinx + c2cosx yp = -y1 ∫ ((y2)(r)/W)dx + y2 ∫ ((y1)(r)/W)dx yp = -sinx ∫ ((cosx)(tan2x)dx) + cosx ∫ ((sinx)(tan2x)dx) yp = -sinx ∫ ((cosx)(sec2x +1)dx) + cosx ∫ ((sinx)(sec2x +1)dx) There's two by parts labeled 1 (left side) and 2 (right side)...

Thank you for the reply. I retried by parts and saw that trig identities simplify the integrals, so I don't have to do by parts a second time. I apologize for not displaying my integrals. My answer is y = c1sinx + c2cos +cosx if you still want to see my work I can display it, but it's a lot...