JEE MAIN - Mathematics (2022 - 25th June Evening Shift - No. 10)
If $$y = y(x)$$ is the solution of the differential equation
$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$ such that $$y(e) = {e \over 3}$$, then y(1) is equal to :
$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$ such that $$y(e) = {e \over 3}$$, then y(1) is equal to :
$${1 \over 3}$$
$${2 \over 3}$$
$${3 \over 2}$$
3
Explanation
$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$
$$ \Rightarrow 2x(xdy - ydx) + 3{y^3}dx = 0$$
$$ \Rightarrow 2\left( {{{xdy - ydx} \over {{y^2}}}} \right) + 3{{dx} \over x} = 0$$
$$ \Rightarrow - {{2x} \over y} + 3\ln x = C$$
$$\because$$ $$y(e) = {e \over 3} \Rightarrow - 6 + 3 = C \Rightarrow C = - 3$$
Now, at $$x = 1$$, $$ - {2 \over y} + 0 = - 3$$
$$y = {2 \over 3}$$
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