Linear Differential Equation: when x=x(y)

[rygza]

(x+y^2)dy=ydx
rewrote as: dx/dy - x/y = y
Realized I had P(y)x and Q(y) rather than the P(x) and Q(x) from equations where y is a function of x. My problem now is after I multiply by the Integrating factor (-1/y):

-1 - x/(y^2) + 1/y(dx/dy)

I tried to make exact but i don't know the proper variables to use. I used

(partial derivative. M/partial deriv. x) = - 1/(y^2) = (partial N/partial y)

Is this proper? Usually for y=y(x) functions it's (partial M/partial y) but if i use that for this problem it doesn't make the equation exact. OR Am I supposed to rewrite so i have dy/dx (and then use the y=y(x) method)?

 

[jackmell]

How about if I write it as:

xdy+y^2dy=ydx

xdy-ydx=-y^2dy

Ok, we know:

d\left(\frac{y}{x}\right)=\frac{xdy-ydx}{x^2}

So that left side could be written as:

x^2d\left(\frac{y}{x}\right)=-y^2dy

Now what happens if I divide throughout by y^2?

 

[rygza]

How about if I write it as:

xdy+y^2dy=ydx

xdy-ydx=-y^2dy

Ok, we know:

d\left(\frac{y}{x}\right)=\frac{xdy-ydx}{x^2}

So that left side could be written as:

x^2d\left(\frac{y}{x}\right)=-y^2dy

Now what happens if I divide throughout by y^2?

lost me on the d(y/x) = ... part

 

[tiny-tim]

lost me on the d(y/x) = ... part

hi rygza!

i suppose you're happy with d(xy) = xdy + ydx ?

that's the equivalent of (xy)' = xy' + yx'.

ok, now start with (y/x)' = (xy' - yx')/x²,

and you get d(y/x) = (xdy - ydx)/x².