JEE Main 2022 — Differential Equations Question with Solution

Given,

$$\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$$

$$\Rightarrow bxdy+cydy+ady=axdx-bydx+adx$$

$$\Rightarrow bxdy+bydx+(cy+a)dy=(ax+a)dx$$

$$\Rightarrow b(xdy+ydx)+(cy+a)dy=(ax+a)dx$$

$$\Rightarrow bd(xy)+(cy+a)dy=(ax+a)dx$$

Integrating both sides, we get

$$\Rightarrow b\int d(xy)+\int (cy+a)dy=\int (ax+a)dx$$

$$\Rightarrow b.xy+c.\frac{y^2}{2}+ay=\frac{a{x}^2}{2}+ax+k$$

$$\Rightarrow \frac{a{x}^2}{2}-\frac{c{y}^2}{2}-bxy+ax-ay+k=0$$

For equation of circle,

Coefficient of x² = Coefficient of y²

$$\therefore$$ $$\frac{a}{2}=-\frac{c}{2}$$

$$\Rightarrow a=-c$$

And coefficient of $$xy=0$$

$$\therefore$$ $$-b=0$$

$$\Rightarrow b=0$$

$$\therefore$$ Circle equation becomes,

$$\frac{a{x}^2}{2}+\frac{a{y}^2}{2}+ax-ay+k=0$$

$$\Rightarrow x^2+y^2+2x-2y+\frac{2k}{a}=0$$

$$\therefore$$ Center $$=(-g,-f)=(-1,1)=(\alpha ,\beta )$$

$$\therefore$$ $$\alpha =-1$$ and $$\beta =1$$

$$\therefore$$ $$\alpha +2\beta =-1+2\times (1)=1$$