Question

If $(x^3{y}^3+xy)\frac{dy}{dx},y\left(1\right)=0$ then when $y=\frac{1}{2}$ the value of $x^2$ equals

• A. $\frac{4}{3}$
• B. $\frac{3}{4}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

Answer: (A)

$\frac{4}{3}$

From given
$\frac{dy}{dx}=\frac{1}{x^3{y}^3+xy}$
$\frac{dx}{dy}=x^3{y}^3+xy\Rightarrow \frac{1}{x^3}\frac{dy}{dx}-\frac{y}{x^2}=y^3$
$\Rightarrow \frac{dt}{dy}+2yt=-2y^3$ $(t=1/x^2)$
$\Rightarrow t.e^{y^2}=-\int 2y{e}^{y^2}{y}^{2}dy$
$\Rightarrow t^2=-e^{y^2}(y^2-1)+K$
$\Rightarrow 1/x^2=1-y^2+K{e}^{{-y}^2}$
$\Rightarrow 1/x^2=1-y^2(\because K=0fory(1)=0)$
$x^2=4/3$ as $y=1/2$