# Differential equation with integrating factor

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

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

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

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

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

• Thanks, I think that I understand – Peter F. Oct 25 '20 at 13:10

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

• Thank you so much – Peter F. Oct 25 '20 at 13:31
• You're very welcome @PeterF. – Satyendra Oct 25 '20 at 13:34

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

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

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

Let $$u=y-x$$ ,

Then $$y=u+x$$

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

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

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

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

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

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

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

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

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

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

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

• Thank you so much – Peter F. Oct 25 '20 at 13:31