Medium MCQ +4 / -1 PYQ · JEE Mains 2021

If the solution curve of the differential equation (2x $-$ 10y3)dy + ydx = 0, passes through the points (0, 1) and (2, $\beta$), then $\beta$ is a root of the equation :

  1. A y<sup>5</sup> $-$ 2y $-$ 2 = 0
  2. B 2y<sup>5</sup> $-$ 2y $-$ 1 = 0
  3. C 2y<sup>5</sup> $-$ y<sup>2</sup> $-$ 2 = 0
  4. D y<sup>5</sup> $-$ y<sup>2</sup> $-$ 1 = 0 Correct answer

Solution

$(2x - 10{y^3})dy + ydx = 0$<br><br>$\Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}$<br><br>$I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}$<br><br>Solution of D.E. is<br><br>$\therefore$ $x\,.\,y = \int {(10{y^2}){y^2}.\,dy}$<br><br>$x{y^2} = {{10{y^5}} \over 5} + C \Rightarrow x{y^2} = 2{y^5} + C$<br><br>It passes through (0, 1) $\to$ 0 = 2 + C $\Rightarrow$ C = $-$2<br><br>$\therefore$ Curve is $x{y^2} = 2{y^5} - 2$<br><br>Now, it passes through (2, $\beta$)<br><br>$2{\beta ^2} = 2{\beta ^5} - 2 \Rightarrow {\beta ^5} - {\beta ^2} - 1 = 0$<br><br>$\therefore$ $\beta$ is root of an equation ${y^5} - {y^2} - 1 = 0$

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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