Differential Equation: The differential equation satisfied by the system of parabolas… | JEE Main 2021

JEE Main 2021 (Online) 18th March Morning ShiftQuestion 1

The differential equation satisfied by the system of parabolas $y^{2}=4a(x+a)$ is :

Differential Equation

$y\left(\frac{dy}{dx}\right)^{2}+2x\left(\frac{dy}{dx}\right)-y=0$✓ Correct answer$y\left(\frac{dy}{dx}\right)+2x\left(\frac{dy}{dx}\right)-y=0$$y\left(\frac{dy}{dx}\right)^{2}-2x\left(\frac{dy}{dx}\right)+y=0$$y\left(\frac{dy}{dx}\right)^{2}-2x\left(\frac{dy}{dx}\right)-y=0$

Solution

Concept. One parameter $a$, so differentiate once and eliminate $a$. Formulas used. (1) $y^2=4a(x+a)\Rightarrow2yy'=4a$Step 1 - Find $a$. By (1), $a=\dfrac{yy'}{2}$. Step 2 - Substitute. $$y^2=2yy'\left(x+\frac{yy'}{2}\right)=2xyy'+y^2y'^2.$$ Step 3 - Divide by $y$. $y=2xy'+y\,y'^2$, i.e. $$y\left(\frac{dy}{dx}\right)^2+2x\frac{dy}{dx}-y=0.$$ Answer: (A)

Differential Equation - JEE Mathematics question (JEE Main 2021 (Online) 18th March Morning Shift)

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