If the curve, y = y (x) represented by the solution of the differential equation (2x y ^ 2 - y)dx + xdy = 0 , passes through the intersection of the lines, 2x - 3y = 1 and 3x + 2y = 8 , then | y(1) | is equal to ___ .

Correct answer: 0 (2x y ^ 2 - y)dx + xdy = 0 2x y ^ 2 dx - ydx + xdy = 0 2xdx = ydx - xdy y ^ 2 = d((x)/(y)) Now integrate x ^ 2 = (x)/(y) + c Now point of intersection of lines are (2, 1) . 4 = (2)/(1) + c ⇒ c = 2 x ^ 2 = (x)/(y) + 2 Now y(1) = - 1 ⇒ | y(1) | = 1

JEE Main 202125 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2021 · 25 February · Evening Shift · Mathematics · Differential Equations

If the curve, y = y (x) represented by the solution of the...

If the curve,

y = y \left(x\right) represented by

the

solution of the differential equation \left(2x{y}^{2} - y\right)dx + xdy = 0

, passes through the intersection of the lines,

2x - 3y = 1 and 3x + 2y = 8

, then

| y\left(1\right) |

is equal to ___ .

Correct answer

Step-by-step solution

\left(2x{y}^{2} - y\right)dx + xdy = 0 $$ 2x{y}^{2}dx - ydx + xdy = 0 $$ 2xdx = \dfrac{ydx - xdy}{{y}^{2}} = d\left(\dfrac{x}{y}\right) $Now integrate$ {x}^{2} = \dfrac{x}{y} + c $Now point of intersection of lines are \left(2, 1\right)$ .

$4 = \dfrac{2}{1} + c ⇒ c = 2$

${x}^{2} = \dfrac{x}{y} + 2$

Now $y\left(1\right) = - 1$

$⇒ | y\left(1\right) | = 1$

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