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QJEEM19Maf61a: JEE Main 2019 Mathematics - Class 12 JEEM19Maf61a Solution: Consider the differential equation, , If value of y is 1 when x …

Consider the differential equation, y2dx+(x−1y)dy=0{y^2}dx + \left( {x - {1 \over y}} \right)dy = 0 , If value of y is 1 when x = 1, then the value of x for which y = 2, is :

Multiple Choice Options

  • A 32−1e{3 \over 2} - {1 \over {\sqrt e }}
  • B 12+1e{1 \over 2} + {1 \over {\sqrt e }}
  • C 52+1e{5 \over 2} + {1 \over {\sqrt e }}
  • D 32−e{3 \over 2} - \sqrt e

Step-by-Step Verified Solution

y2dx+(x−1y)dy=0{y^2}dx + \left( {x - {1 \over y}} \right)dy = 0

⇒dxdy+xy2=1y3\Rightarrow {{dx} \over {dy}} + {x \over {{y^2}}} = {1 \over {{y^3}}}

Integrating factor (I.F) = e−1y{e^{ - {1 \over y}}}

Now x. e−1y{e^{ - {1 \over y}}} = ∫e−1y1y3dy\int {{e^{ - {1 \over y}}}{1 \over {{y^3}}}dy} by putting −1y{ - {1 \over y}} = t

x.et = ∫et(−t)dt\int {{e^t}( - t)dt}

⇒xet=−(t.et−et)+c\Rightarrow x{e^t} = - (t.{e^t} - {e^t}) + c

⇒xe−1y=e−1y(1+1y)+c\Rightarrow x{e^{ - {1 \over y}}} = {e^{ - {1 \over y}}}\left( {1 + {1 \over y}} \right) + c

⇒x=1+1y+c.e1y\Rightarrow x = 1 + {1 \over y} + c.e^{1 \over y}

It passes through (1, 1)

∴\therefore c = −1e-{1 \over e}

Equation of the curve is

x=1+1y−e1y−1x = 1 + {1 \over y} - {e^{{1 \over y} - 1}} It passes through (k, 2)

∴\therefore k = 1+12−e12−11 + {1 \over 2} - {e^{{1 \over 2} - 1}} = 323 \over 2 - 1e1 \over{\sqrt e}

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