If y = y(x) is the solution of the differential equation [2x^2dy dx - 2xy + 3y^2 = 0] such…

Question

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 :

Accepted Answer

$$2{x^2}{{dy} \over {dx}} - 2xy + 3{y^2} = 0$$

$$ \Rightarrow 2x(xdy - ydx) + 3{y^3}dx = 0$$

$$ \Rightarrow 2\left( {{{xdy - ydx} \over {{y^2}}}} \right) + 3{{dx} \over x} = 0$$

$$ \Rightarrow  - {{2x} \over y} + 3\ln x = C$$

$$\because$$ $$y(e) = {e \over 3} \Rightarrow  - 6 + 3 = C \Rightarrow C =  - 3$$

Now, at $x = 1$, $$ - {2 \over y} + 0 =  - 3$$

$$y = {2 \over 3}$$

Solutions · 1