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 :
Solution
<p>$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$</p>
<p>$\Rightarrow 2x(xdy - ydx) + 3{y^3}dx = 0$</p>
<p>$\Rightarrow 2\left( {{{xdy - ydx} \over {{y^2}}}} \right) + 3{{dx} \over x} = 0$</p>
<p>$\Rightarrow - {{2x} \over y} + 3\ln x = C$</p>
<p>$\because$ $y(e) = {e \over 3} \Rightarrow - 6 + 3 = C \Rightarrow C = - 3$</p>
<p>Now, at $x = 1$, $- {2 \over y} + 0 = - 3$</p>
<p>$y = {2 \over 3}$</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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