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QJEEM21Ma930e: JEE Main 2021 Mathematics - Class 12 JEEM21Ma930e Solution: If y = y(x) is the solution curve of the differential equation…

If y = y(x) is the solution curve of the differential equation x2dy+(y−1x)dx=0{x^2}dy + \left( {y - {1 \over x}} \right)dx = 0 ; x > 0 and y(1) = 1, then y(12)y\left( {{1 \over 2}} \right) is equal to :

Multiple Choice Options

  • A 32−1e{3 \over 2} - {1 \over {\sqrt e }}
  • B 3+1e3 + {1 \over {\sqrt e }}
  • C 3 + e
  • D 3 −- e

Step-by-Step Verified Solution

x2dy+(y−1x)dx=0{x^2}dy + \left( {y - {1 \over x}} \right)dx = 0 ; x > 0, y(1) = 1

x2dy+(xy−1)xdx=0{x^2}dy + {{(xy - 1)} \over x}dx = 0

x2dy=(xy−1)xdx{x^2}dy = {{(xy - 1)} \over x}dx

dydx=1−xyx3{{dy} \over {dx}} = {{1 - xy} \over {{x^3}}}

dydx=1x3−yx2{{dy} \over {dx}} = {1 \over {{x^3}}} - {y \over {{x^2}}}

dydx=1x2.y=1x3{{dy} \over {dx}} = {1 \over {{x^2}}}.y = {1 \over {{x^3}}}

If e∫1x2dx=e−1x{e^{\int {{1 \over {{x^2}}}dx} }} = {e^{ - {1 \over x}}}

ye−1x=∫1x3.e−1xy{e^{ - {1 \over x}}} = \int {{1 \over {{x^3}}}.{e^{ - {1 \over x}}}}

ye−1x=e−x(1+1x)+Cy{e^{ - {1 \over x}}} = {e^{ - x}}\left( {1 + {1 \over x}} \right) + C

1. e−1=e−1(2)+C1.\,{e^{ - 1}} = {e^{ - 1}}(2) + C

C=−e−1=−1eC = - {e^{ - 1}} = - {1 \over e}

ye−1x=e−1x(1+1x)−1ey{e^{ - {1 \over x}}} = {e^{ - {1 \over x}}}\left( {1 + {1 \over x}} \right) - {1 \over e}

y(12)=3−1e×e2y\left( {{1 \over 2}} \right) = 3 - {1 \over e} \times {e^2}

y(12)=3−ey\left( {{1 \over 2}} \right) = 3 - e

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