Question

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
$$y-x\frac{dy}{dx}=y^2+\frac{dy}{dx}$$

Answer 1

We have,
$$y-x\frac{dy}{dx}=y^2+\frac{dy}{dx}$$
$$\Rightarrow \frac{dy}{dx}(x+1)=y(1-y)$$
$$\Rightarrow \frac{dy}{y(1-y)}=\frac{dx}{(x+1)}$$
Integrating both sides, we get
$$\int \frac{dy}{y(1-y)}=\int \frac{dx}{x+1}$$
$$\Rightarrow \int (\frac{1}{y}+\frac{1}{1-y})dy=\int \frac{dx}{x+1}$$
$$\Rightarrow \log \left|y\right|-\log |1-y|=\log |x+1|+C.....\left(1\right)$$
$$\text{ Since the curve passes throught the point }(2,2),\text{ it satisfies the equation of the curve . }$$
$$\Rightarrow \log \left|2\right|-\log |1-2|=\log |2+1|+C$$
$$\Rightarrow C=\log \left|\frac{2}{3}\right|$$
$$\text{ Putting the value of C in }\left(1\right),\text{ we get }$$
$$\log \left|y\right|-\log |1-y|=\log |x+1|+\log \left|\frac{2}{3}\right|$$
$$\Rightarrow \log \left|\frac{y}{(1-y)}\right|=\log \left|\frac{2(x+1)}{3}\right|$$
$$\Rightarrow \left|\frac{y}{(1-y)}\right|=\left|\frac{2(x+1)}{3}\right|$$
$$\Rightarrow \frac{y}{(1-y)}=\pm \frac{2(x+1)}{3}$$
$$\Rightarrow \frac{y}{(1-y)}=\frac{2(x+1)}{3}or\frac{y}{(1-y)}=-\frac{2(x+1)}{3}$$
$$\text{ Here, given point }(2,2)\text{ does not satisfy }\frac{y}{(1-y)}=\frac{2(x+1)}{3}$$
$$\text{ But it satisfy }\frac{y}{(1-y)}=-\frac{2(x+1)}{3}$$
$$\therefore \frac{y}{(1-y)}=-\frac{2(x+1)}{3}$$
$$\Rightarrow \frac{y}{(y-1)}=\frac{2(x+1)}{3}$$
$$\Rightarrow 3y=2(x+1)(y-1)$$
$$\Rightarrow 3y=2xy-2x+2y-2$$
$$\Rightarrow 2xy-2x-y-2=0$$