QJEEM20Ma87e0: JEE Main 2020 Mathematics - Class 12 JEEM20Ma87e0 Solution: If a curve y = f(x), passing through the point (1, 2), is the…

Question

Verified Solution • 2025–26 Session moderate

If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x²dy= (2xy + y²)dx, then

f\left( {{1 \over 2}} \right)

is equal to :

Accepted Answer

2{x^2}dy = \left( {2xy + {y^2}} \right)dx

\Rightarrow 2{x^2}{{dy} \over {dx}} = 2xy + {y^2}

\Rightarrow {{2{x^2}} \over {2{x^2}{y^2}}}{{dy} \over {dx}} = {{2xy} \over {2{x^2}{y^2}}} + {{{y^2}} \over {2{x^2}{y^2}}}

\Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over x}{1 \over y} = {1 \over {2{x^2}}}

Let

- {1 \over y} = t

\Rightarrow {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}

\Rightarrow {{dt} \over {dx}} + t{1 \over x} = {1 \over {2{x^2}}}

This is linear differentiatial equation.

\therefore I.F = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}}

\therefore t.{e^{\ln x}} = \int {{1 \over {2{x^2}}}.{e^{\ln x}}dx}

\Rightarrow - {1 \over y}.x = \int {{1 \over {2{x^2}}}.xdx}

\Rightarrow - {x \over y} = {1 \over 2}\int {{{dx} \over x}}

\Rightarrow - {x \over y} = {1 \over 2}\ln x + c

This curve passes through the point (1, 2)

\therefore - {1 \over 2} = 0 + c

\Rightarrow c = - {1 \over 2}

\therefore - {x \over y} = {1 \over 2}\ln x - {1 \over 2}

\Rightarrow - {{2x} \over y} = \ln x - 1

\Rightarrow y = {{2x} \over {1 - \ln x}}

\Rightarrow f\left( x \right) = {{2x} \over {1 - \ln x}}

So,

f\left( {{1 \over 2}} \right) = {{2 \times {1 \over 2}} \over {1 - \ln \left( {{1 \over 2}} \right)}} = {1 \over {1 + {{\log }_e}2}}

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