Question

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

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

Answer 1

Answer: (B)

$x^{3}dy+xydx=x^{2}dy+2ydx$
$\Rightarrow dy\left(x^3-x^2\right)=dx(2y-xy)$
$\Rightarrow -\int \frac{1}{y}dy=\int \frac{x-2}{x^2(x-1)}dx$
$\Rightarrow -\ell ny=\int \left(\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x-1)}\right)dx$
Where $A=1,B=+2,C=-1$
$\Rightarrow -\ell ny=\ln x-\frac{2}{x}-\ln (x-1)+\lambda$
$\Rightarrow y(2)=e$
$\Rightarrow -1=\ln 2-1-0+\lambda$
$\therefore \lambda =-\ln 2$
$\Rightarrow \ell ny=-\ln x+\frac{2}{x}+\ln (x-1)+\ln 2$
Now put $x=4$ in equation
$\Rightarrow \ell ny=-\ln 4+\frac{1}{2}+\ln 3+\ln 2$
$\Rightarrow \ell ny=\ln \left(\frac{3}{2}\right)+\frac{1}{2}\ell ne$
$\Rightarrow y=\frac{3}{2}\sqrt{e}$