Solution of linear differential equation

[clope023]

Homework Statement

solve differential equation

xdy + (xy+y-1)dx=0

Homework Equations

dy/dx + P(x)y = Q(x)

u = exp[int(P(x)] - integrating factor

exp[int(P(x))]*[dy/dx+P(x)y = Q(x)]

=> d/dx[P(x)]y = exp[int(P(x))]Q(x)

int[d/dx[P(x)]y = exp[int(P(x))]Q(x)]

solution is with respect to y.

The Attempt at a Solution

xdy = -(xy+y-1)dx

dy/dx = (1-y+xy)/x

dx/dy = x/(1-y+xy)

dx/dy = x-x/y+1/y

dx/dy - x + x/y = 1/y

I'm stuck here, I'm just not sure who to get the equation into a linear form where I could take the integrating factor and solve for x in the case since I switched the dependent variables to see if that wouldn't make it more complicated, any help from that point on would be greatly appreciated.

 

[th3plan]

divide the whole equation by dx, then u get x(dy/dx)+xy+y-1=0

move the 1 to other side, and factor a y out , and u get

x(dy/dx)+y(x+1)=1

divide by x

(dy/dx)+[(x+1)/x]y=1/2

now u in form ,and u have P(x)Y :)

the rest i think you know how to do it