Linear Differential Equation: when x=x(y)

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(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)?
 
Last edited:
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How about if I write it as:

[tex]xdy+y^2dy=ydx[/tex]

[tex]xdy-ydx=-y^2dy[/tex]

Ok, we know:

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

So that left side could be written as:

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

Now what happens if I divide throughout by y^2?
 
38
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How about if I write it as:

[tex]xdy+y^2dy=ydx[/tex]

[tex]xdy-ydx=-y^2dy[/tex]

Ok, we know:

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

So that left side could be written as:

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

Now what happens if I divide throughout by y^2?
lost me on the d(y/x) = ... part
 

tiny-tim

Science Advisor
Homework Helper
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242
lost me on the d(y/x) = ... part
hi rygza! :smile:

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.
 

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