Question

If y = y(x) is the solution of the differential equation

$2x^2\frac{dy}{dx}-2xy+3y^2=0$ such that $y(e)=\frac{e}{3},$

then y(1) is equal to

• A. ⅓
• B. ⅔
• C. 3/2
• D. 3

Answer 1

Answer: (A)

⅓

Answer 2

The correct option is(B): $\frac{2}{3}$.

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

$⇒2x(xdy-ydx)+3y^2dx=0$

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

$∵y(e)=\frac{e}{3}⇒-6+3=c⇒c=-3$

Now,at

$x=1,-\frac{2}{y}+0=-3$

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