"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 _________."

Question

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

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

$\Rightarrow y(1) = - 1$

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

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 _________.

Solution

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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 _________.

Solution

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More Differential Equations questions

Drill 25 more like these. Every day. Free.

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 _________.