QJEEM21Ma6b37: JEE Main 2021 Mathematics - Class 12 JEEM21Ma6b37 Solution: If the curve, y = y(x) represented by the solution of the…

Question

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If the curve, y = y(x) represented by the solution of the differential equation (2xy²

-

y)dx + xdy = 0, passes through the intersection of the lines, 2x

-

3y = 1 and 3x + 2y = 8, then |y(1)| is equal to _________.

Accepted Answer

Given,

(2x{y^2} - y)dx + xdx = 0

\Rightarrow {{dy} \over {dx}} + 2{y^2} - {y \over x} = 0

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

{1 \over y} = z

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

\Rightarrow {{dz} \over {dx}} + z\left( {{1 \over x}} \right) = 2

I. F.

= {e^{\int {{1 \over x}dx} }} = x

\therefore

z(x) = \int {2(x)dx} = {x^2} + c

\Rightarrow {x \over y} = {x^2} + c

As it passes through P(2, 1)

[Point of intersection of

2x - 3y = 1

and

3x + 2y = 8

]

\therefore

{2 \over 1} = 4 + c

\Rightarrow c = - 2

\Rightarrow {x \over y} = {x^2} - 2

Put x = 1

{1 \over y} = 1 - 2 = - 1

\Rightarrow y(1) = - 1

\Rightarrow |y(1)|\, = 1

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