Differential equation

Question

X(dy/dx) - y=(x^p) +1, where x is a real number and is not 0. p is a positive integer, p>1. Solve the differential equation given that y=-1 when x=1. Give ans in form y=f(x)

Answer 1

(x(dy/dx) - y = x^p +1)÷x puts original equation in form (dy/dx) + yP(x) =Q(x) and gives dy/dx - y(1/x) = x^p-1 + 1/x.

P(x) here is -(1/x) or -x^-1 and ∫P(x)dx is -ln x. Then e^∫P(x)dx equal to e^{∫‾ln x} gives 1/x as an integrating factor. The general solution or "primitive" is given by ye^∫P(x)dx = ∫Q(x)e^∫P(x)dxdx +C. For this problem, the general solution is y(1/x) =∫(x^p-1 + x^-1)(1/x)dx + C which is rewritten as y(1/x) = ∫(x^p-2 + x^-2)dx + C which gives yx^-1 = (p-1)^-1x^p-1 - x^-1 + C. This last equation is multiplied by x to give y = (p - 1)^-1x^p -1 +Cx. When y = -1 and x = 1, y = (p - 1)^-1x^p -1 + Cx becomes -1 = (p - 1)^-1(1)^p -1 +C or (p - 1)^-1 + C = 0 which gives C = (1 - p)^-1. Then, for (x,y) = (1,-1), y = (p - 1)^-1x^p - 1 + x(1 - p)^-1 equal to (x^p - x - p + 1) ÷ (p - 1). AS A CHECK: rewrite x(dy/dx) - y = x^p + 1 using y = (p - 1)^-1x^p - 1 + Cx; x((p - 1)^-1px^{p - 1} -0 + C) - ((p - 1)^-1x^p -1 + Cx) gives (p - 1)^-1(p - 1)x^p +Cx + 1 - Cx which simplifies to x^p + 1.