Recent content by Jen_Jer_888

After I get y = -z3 - (2/4)z4 - (2/5)z5 - (2/6)z6 - (2/7)z7 + ... do I need to plug (x+1) back in and get y in terms of x? Can I replace x + 1 with x now and get y = -x3 - (2/4)x4 - (2/5)x5 - (2/6)x6 - (2/7)x7 + ... If not, expanding each of those with x + 1 still in there would be a huge...

Thank you for the response! Here is my updated attempt. Can you verify it or explain if I go wrong? Let z = x+1. dy/dx = 2/x + 3 - x2 => x(dy/dx) = 2+ 3x - x3 => -x(dy/dx) = x3 - 3x - 2. This factors giving -x(dy/dx) = (x+1)2(x-2) So dy/dx = [(x+1)2(x-2)]/(-x). So dy/dx = [z2(z-3)]/(-(z-1))...

Homework Statement Solve the fluxional equation (y with a dot on top)/(x with a dot on top) = 2/x + 3 - x^2 by first replacing x by (x + 1) and then using power series techniques.Homework Equations dy/dx = 2/x + 3 - x^2 The Attempt at a Solution First, I believe the fluxional (y with a dot...

I know how to do problems like y' + y = 0 where you can replace y' and y with a series in sigma notation, manipulate and compare coefficients. But how do you solve a differential by power series that does not also include y or a higher order derivative? For example, y' = -(x^2) + 2/x + 3...