JEE Main 2021 (Online) 27th August Morning ShiftQuestion 1
Let us consider a curve, $y=f(x)$ passing through the point $(-2,2)$ and the slope of the tangent to the curve at any point $(x,f(x))$ is given by $f(x)+xf'(x)=x^{2}$. Then :
Differential Equation
Solution
Concept. $f+xf'$ is the derivative of $xf$. Formulas used. (1) $\dfrac{d}{dx}\big(x f(x)\big)=f(x)+xf'(x)$Step 1 - Integrate. By (1), $\big(xf\big)'=x^2$, so $xf(x)=\dfrac{x^3}{3}+C$. Step 2 - Through $(-2,2)$. $-4=-\frac83+C$, so $C=-\frac43$. Step 3 - The curve. $3xf(x)=x^3-4$, i.e. $x^3-3xf(x)-4=0$. Answer: (A)