Let x = x(y) be the solution of the differential equation 2y^2 (dx)/(dy) - 2xy + x^2 = 0, …

Question

Let $x = x(y)$ be the solution of the differential equation $2y^2 \frac{dx}{dy} - 2xy + x^2 = 0$, $y > 1$, $x(e) = e$.Then $x(e^2)$ is equal to:

Accepted Answer

$2y(ydx-xdy)+{x}^{2}dy=0$

$\Rightarrow -2y{x}^{2}\, d(\frac{y}{x})+{x}^{2}dy=0$

$\Rightarrow -2y\, d(\frac{y}{x})+dy=0$

$\Rightarrow -2d(\frac{y}{x})+\frac{1}{y}dy=0$

$\Rightarrow -\frac{2y}{x}+{\log }_{e}y=C$

Given $x(e)=e$

$\Rightarrow C=-1$

$	herefore -\frac{2y}{x}+{\log }_{e}y=-1$

$\Rightarrow \frac{2y}{x}-{\log }_{e}y=1$

Put $y={e}^{2}$, we get

$\Rightarrow \frac{2{e}^{2}}{x}-2=1	o \frac{2{e}^{2}}{x}=3	o x=\frac{2{e}^{2}}{3}$

$\Rightarrow x({e}^{2})=\frac{2{e}^{2}}{3}$

Solutions · 1