Verified Solution • 2025–26 Session moderate

QJEEM21Ma72d7: JEE Main 2021 Mathematics - Class 12 JEEM21Ma72d7 Solution: If the solution curve of the differential equation (2x 10y 3…

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

Multiple Choice Options

  • A y5 −- 2y −- 2 = 0
  • B 2y5 −- 2y −- 1 = 0
  • C 2y5 −- y2 −- 2 = 0
  • D y5 −- y2 −- 1 = 0

Step-by-Step Verified Solution

(2x−10y3)dy+ydx=0(2x - 10{y^3})dy + ydx = 0

⇒dxdy+(2y)x=10y2\Rightarrow {{dx} \over {dy}} + \left( {{2 \over y}} \right)x = 10{y^2}

I.F.=e∫2ydy=e2ln⁡(y)=y2I.F. = {e^{\int {{2 \over y}dy} }} = {e^{2\ln (y)}} = {y^2}

Solution of D.E. is

∴\therefore x . y=∫(10y2)y2. dyx\,.\,y = \int {(10{y^2}){y^2}.\,dy}

xy2=10y55+C⇒xy2=2y5+Cx{y^2} = {{10{y^5}} \over 5} + C \Rightarrow x{y^2} = 2{y^5} + C

It passes through (0, 1) →\to 0 = 2 + C ⇒\Rightarrow C = −- 2

∴\therefore Curve is xy2=2y5−2x{y^2} = 2{y^5} - 2

Now, it passes through (2, β\beta )

2β2=2β5−2⇒β5−β2−1=02{\beta ^2} = 2{\beta ^5} - 2 \Rightarrow {\beta ^5} - {\beta ^2} - 1 = 0

∴\therefore β\beta is root of an equation y5−y2−1=0{y^5} - {y^2} - 1 = 0

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