JEE Main 2016 (Offline)Question 1
If a curve $y=f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation $y(1+xy)\,dx=x\,dy$, then $f\left(-\frac12\right)$ is equal to
Differential Equation
Solution
Concept. $y'=\frac yx+y^2$ is a Bernoulli equation; $u=\frac1y$ makes it linear. Formulas used. (1) $u=\frac1y\Rightarrow u'=-\dfrac{y'}{y^2}$Step 1 - Linear form. By (1), $u'=-\frac ux-1$, so $(xu)'=-x$ and $xu=-\frac{x^2}{2}+C$. Step 2 - Through $(1,-1)$. $u=-1$: $-1=-\frac12+C$, so $C=-\frac12$. Then $u=-\dfrac{x^2+1}{2x}$, so $$y=-\frac{2x}{x^2+1}.$$ Step 3 - Value. $f\left(-\frac12\right)=\dfrac{1}{\frac14+1}=\dfrac45$. Answer: (B)