Question

Let $y=y(x)$ be the solution of the differential equation $\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1$.Then $6 y^2( e )$ is equal to

• A. $e^2$
• B. $\frac{3}{2} e^2$
• C. $3 e^2$
• D. $2 e ^2$

Answer 1

Answer: (D)

$2 e ^2$

Answer 2

The correct answer is (D) : $2 e ^2$
(x2−3y2)dx+3xydy=0
dxdy​=3xy3y2−x2​⇒dxdy​=xy​−31​yx​.....(1)
Put y=vx
dxdy​=v+xdxdv​
(1) ⇒v+xdxdv​=v−31​v1​
⇒vdv=3x−1​
Integrating both side
2v2​=3−1​lnx+c
⇒2x2y2​=3−1​lnx+c
y(1)=1
⇒21​=c
⇒2x2y2​=3−1​lnx+21​
⇒y2=−32​x2lnx+x2
y2(e)=−32​e2+e2=3e2​
⇒6y2(e)=2e2