Question

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

• 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 $xy=6$

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

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

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

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

Let us denote $y\left(x\right)$ as $y$

$\Rightarrow xy=x^2{y}^{\prime }+2xy$

$\Rightarrow x^2{y}^{\prime }=xy-2xy$

$\Rightarrow x^2{y}^{\prime }=-xy$

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

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

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

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

$\Rightarrow \ln \left|xy\right|=c$

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

$\ln |2\times 3|=c$

$\Rightarrow c=\ln \left|6\right|$

$\Rightarrow \ln \left|xy\right|=\ln \left|6\right|$

$\Rightarrow xy=6$