JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 25)
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 _________.
Answer
1
Explanation
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
$${1 \over y} = 1 - 2 = - 1$$
$$ \Rightarrow y(1) = - 1$$
$$ \Rightarrow |y(1)|\, = 1$$
$$(2x{y^2} - y)dx + xdx = 0$$
$$ \Rightarrow {{dy} \over {dx}} + 2{y^2} - {y \over x} = 0$$
$$ \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} + {1 \over y}\left( {{1 \over x}} \right) = 2$$
$${1 \over y} = z$$
$$ - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dz} \over {dx}}$$
$$ \Rightarrow {{dz} \over {dx}} + z\left( {{1 \over x}} \right) = 2$$
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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