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JEE Main 202118 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Differential Equations

JEE Main 2021 · 18 March · Morning Shift · Mathematics · Differential Equations

The differential equation satisfied by the system of parabolas y ^...

The differential equation satisfied by the system of parabolas y2=4a(x+a){y}^{2} = 4a ( x + a ) is

Options

  1. Ay(dydx)2−2x(dydx)−y=0y{\left(\dfrac{dy}{dx}\right)}^{2} - 2x\left(\dfrac{dy}{dx}\right) - y = 0
  2. By(dydx)2−2x(dydx)+y=0y{\left(\dfrac{dy}{dx}\right)}^{2} - 2x\left(\dfrac{dy}{dx}\right) + y = 0
  3. Cy(dydx)2+2x(dydx)−y=0y{\left(\dfrac{dy}{dx}\right)}^{2} + 2x\left(\dfrac{dy}{dx}\right) - y = 0
  4. Dy(dydx)+2x(dydx)−y=0y\left(\dfrac{dy}{dx}\right) + 2x\left(\dfrac{dy}{dx}\right) - y = 0

Correct answer

C. y(dydx)2+2x(dydx)−y=0y{\left(\dfrac{dy}{dx}\right)}^{2} + 2x\left(\dfrac{dy}{dx}\right) - y = 0

Step-by-step solution

y2=4ax+4a2{y}^{2} = 4ax + 4{a}^{2}

differentiate with respect to xx

⇒2ydydx=4a⇒ 2y\dfrac{dy}{dx} = 4a

⇒a=(y2dydx)⇒ a = \left(\dfrac{y}{2}\dfrac{dy}{dx}\right)

so, required differential equation is

y2=(4×y2dydx)x+4(y2dydx)2{y}^{2} = \left(4 \times \dfrac{y}{2}\dfrac{dy}{dx}\right)x + 4{\left(\dfrac{y}{2}\dfrac{dy}{dx}\right)}^{2}

⇒y2(dydx)2+2xy(dydx)−y2=0⇒ {y}^{2}{\left(\dfrac{dy}{dx}\right)}^{2} + 2xy\left(\dfrac{dy}{dx}\right) - {y}^{2} = 0

⇒y(dydx)2+2x(dydx)−y=0⇒ y{\left(\dfrac{dy}{dx}\right)}^{2} + 2x\left(\dfrac{dy}{dx}\right) - y = 0

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