JEE Main 2021 (Online) 25th February Morning ShiftQuestion 1
If a curve passes through the origin and the slope of the tangent to it at any point $(x,y)$ is $\frac{x^{2}-4x+y+8}{x-2}$, then this curve also passes through the point :
Differential Equation
Solution
Concept. Shift to $t=x-2$; then $x^2-4x+8=t^2+4$ and the equation is linear with IF $\frac1t$. Formulas used. (1) $\left(\dfrac yt\right)'=\dfrac{ty'-y}{t^2}$Step 1 - Shift. With $t=x-2$: $$\frac{dy}{dt}=\frac{t^2+4+y}{t}\ \Longrightarrow\ t\,y'-y=t^2+4.$$ Step 2 - Integrate. By (1), $\left(\dfrac yt\right)'=1+\dfrac{4}{t^2}$, so $$\frac yt=t-\frac4t+C\ \Longrightarrow\ y=t^2-4+Ct.$$ Step 3 - Through the origin. $x=0$ means $t=-2$: $0=4-4-2C$, so $C=0$. Hence $y=(x-2)^2-4=x^2-4x$. Step 4 - Check the options. At $x=5$: $y=25-20=5$. So $(5,5)$ lies on it. Answer: (B)