Question

If $x^{3}dy+xydx=x^{2}dy+2ydx;y\left(2\right)=e$ and $x>1$, then $y\left(4\right)$ is equal to:

• A. $\frac{\sqrt{e}}{2}$
• B. $\frac{3}{2}\sqrt{e}$
• C. $\frac{1}{2}+\sqrt{e}$
• D. $\frac{3}{2}+\sqrt{e}$

Answer 1

The correct option is B $\frac{3}{2}\sqrt{e}$

$(x^3-x^2)dy=(2-x)ydx$
$\int \frac{dy}{y}=\int \frac{2-x}{x^2(x-1)}dx$
$\int \frac{dy}{y}=-\int \frac{x-2}{x^2(x-1)}dx$

Let $\frac{x-2}{x^2(x-1)}=\frac{p}{x}+\frac{q}{x^2}+\frac{r}{x-1}$
$\Rightarrow p=1,q=2,r=-1$

$\therefore \int \frac{dy}{y}=-\int \frac{1}{x}dx-\int \frac{2}{x^2}dx+\int \frac{1}{x-1}dx$
$\Rightarrow \ln \left|y\right|=-\ln |x|+\frac{2}{x}+\ln |x-1|+C$
At $x=2\Rightarrow y=e$
$1=1-\ln 2+C$
$\Rightarrow C=\ln 2$
$\Rightarrow \ln \left|y\right|=\frac{2}{x}-\ln \left|x\right|+\ln |x-1|+\ln 2$
At $x=4$
$\ln y=\frac{1}{2}-\ln 4+\ln 3+\ln 2$
$\Rightarrow \ln y=\ln \left(\frac{3}{2}\right)+\frac{1}{2}$
$\Rightarrow y=\frac{3}{2}\cdot e^{\frac{1}{2}}=\frac{3}{2}\sqrt{e}$