Question

A curve passing through (2, 3) and satisfying the differential equation ${\int }_0^{x}ty\left(t\right)dt=x^{2}y\left(x\right),(x>0)$ is

• A. $x^2+y^2=13$
• B. $y^2=\frac{9}{2}x$
• C. $\frac{x^2}{8}+\frac{y}{18}=1$
• D. $xy=6$

Answer 1

The correct option is D $xy=6$

${\int }_0^{x}ty\left(t\right)dt=x^{2}y\left(x\right)$

Differentiating w.r.t $x$, we get

$xy=2xy+x^2\frac{dy}{dx}\Rightarrow x^2\frac{dy}{dx}+xy=0$

$x\frac{dy}{dx}+y=0\Rightarrow xdy+ydx=0$

$d\left(xy\right)=0\Rightarrow xy=c$

$\therefore$ since it passes through (2,3)

$\therefore c=6$

Hence the curve is, $xy=6$