Differential equation with integrating factor

Question

Hey I am supposed to solve the following differential equation:

$(1-x^{2}y)dx+x^{2}(y-x)dy=0$

I found integrating factor:

$\varphi (x)=-\frac{2}{x}$

So I multiply my original equation and I got: $\left ( \frac{1}{x^{2}} -y\right )dx+\left ( y-x \right )dy=0$

But then I try to integrate it and I got stuck. The answer should be:

$y^{2}-2xy-\frac{2}{x}=C$

Can somebody help me?

Thanks

Answer 1

Strating from your last line: $$\left ( \frac{1}{x^{2}} -y\right )dx+\left ( y-x \right )dy=0$$ $$\frac{1}{x^{2}} dx+ y dy-(xdy+ydx)=0$$

$$\frac{1}{x^{2}} dx+ y dy-dxy=0$$ Integration gives us: $$-\frac{2}{x} + y^2 -2xy=c$$ $$\frac{2}{x} - y^2 +2xy=C$$

Answer 2

$P(x,y)dx + Q(x,y)dy = 0$

$P(x,y) = \frac{1}{x^2} - y, Q(x,y) = y - x$

$\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y} = -1$

So this is exact differential equation.

To solve the exact differential equation, let's write two differential equations that define function $u(x,y)$

$\frac{\partial u}{\partial x} = P(x,y)$

$\frac{\partial u}{\partial y} = Q(x,y)$

$u(x,y) = \int P(x,y) dx + \varphi(y) = - \frac{1}{x} - xy + \varphi(y)$ ...(i)

$\varphi'(y) = Q(x,y) - \frac{\partial \, (- \frac{1}{x} - xy)}{\partial y} = y - x + x = y$

$\varphi(y) = \frac{y^2}{2}$ ...(ii)

Substituting (ii) in (i),

$u(x,y) = \frac{y^2}{2} - xy - \frac{1}{x} = C_1$

Or, $y^2 - 2xy - \frac{2}{x} = C$

Answer 3

$(1-x^2y)~dx+x^2(y-x)~dy=0$

$x^2(y-x)~dy=(x^2y-1)~dx$

$(y-x)\dfrac{dy}{dx}=y-\dfrac{1}{x^2}$

Let $u=y-x$ ,

Then $y=u+x$

$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$

$\therefore u\left(\dfrac{du}{dx}+1\right)=u+x-\dfrac{1}{x^2}$

$u\dfrac{du}{dx}+u=u+x-\dfrac{1}{x^2}$

$u\dfrac{du}{dx}=x-\dfrac{1}{x^2}$

$u~du=\left(x-\dfrac{1}{x^2}\right)$

$\int u~du=\int\left(x-\dfrac{1}{x^2}\right)~dx$

$\dfrac{u^2}{2}=\dfrac{x^2}{2}+\dfrac{1}{x}+c$

$u^2=x^2+\dfrac{2}{x}+C$

$(y-x)^2=x^2+\dfrac{2}{x}+C$

$y^2-2xy+x^2=x^2+\dfrac{2}{x}+C$

$y^2-2xy=\dfrac{2}{x}+C$