Homogenous ordinary equation - Homogeneous

Question

The question is:

$(x-y)dx + xdy = 0$

Trying to solve:

$ \\M(x,y) = (x-y) \\N(x,y) = x $

$ \\Kx - Ky = K(x-y) \Rightarrow \text{ Homogeneous} \\Kx = K(x) \Rightarrow \text{Homogeneous}$

$ \\y = vx \\dy = vdx+xdv$

$ \\(x-vx)dx+x(vdx+xdv)=0 \\xdx + x^2dv = 0 $

I'm stucked here. I know I have to integrate now. The answer is

$$x=e^{-y/x}+c$$ but I can't get there.

What I did is:

$$ \\x^2dv=-xdx \\\int dv=-\int\frac{x}{x^2}dx \\\int dv=-\int\frac{1}{x}dx \\v+c_1=-ln(x)+c_2 \\\frac{y}{x}=-ln(x)+c_2-c_1 \\-x(ln(x)+c_2-c_1)=y $$

And now? What should I do?

Answer 1

I'm a little confused by your notation earlier so I haven't checked it through, but at the end it's simple enough to rearrange your last line by dividing by $-x$ and then exponentiating. (Let $c=c_1-c_2$.) You get almost what you were looking for. (See my comment.)

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Exponentiating means raising $e$ to the powers given. $$a=b \implies e^a=e^b$$ Note that $e^{\ln y}=y$.

Answer 2

$\mathbf{HINT}$: Try the following:$$\frac{xdx}{x^2}=\frac{ydx-xdy}{x^2}=-d\left(\frac{y}{x}\right)$$