Question

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

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

Answer 1

The correct option is D $xy=6$

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

Differentiating both sides with respect to $x,$and applying Lebnitz rule, we get,

$x.y\left(x\right)\frac{d}{dx}\left(x\right)-0=x^2{y}^{\prime }\left(x\right)+y\left(x\right)\frac{d}{dx}\left(x^2\right)$

$\Rightarrow xy\left(x\right)=x^2.y^{\prime }\left(x\right)+2x.y\left(x\right)$

We will denote $y\left(x\right)$ as $y$

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

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

$\Rightarrow x\frac{dy}{dx}=-y$

$\Rightarrow \frac{dy}{y}=-\frac{dx}{x}$

$\Rightarrow \ln \left|y\right|=-\ln \left|x\right|+c$

$\Rightarrow \ln \left|y\right|+\ln \left|x\right|=c$

$\Rightarrow \ln \left(xy\right)=c$

Now given that $(2,3)$ lies on the curve, hence

$\Rightarrow \ln (2\times 3)=c$

$\Rightarrow c=ln\left(6\right)$

$\Rightarrow \ln \left(xy\right)=\ln \left(6\right)$

$\therefore xy=6$