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

$\frac{3}{2}\sqrt{e}$

Explanation for the correct option.

Step 1: Find the integral of the given equation.

To find the value of

Step 2: Find the value of $c$

Now, we have $y\left(2\right)=e$, that means $x=2,\mathrm{and}y=e$.

So, $\begin{array}{rcl}\ln y& =& -\ln x+\frac{2}{x}+\ln (x-1)+c\end{array}$ will be

$\begin{array}{rcl}\ln e& =& -\ln 2+\frac{2}{2}+\ln \left(2-1\right)+c\\ & \Rightarrow & 1=-\ln 2+1+0+c\\ & \Rightarrow & c=\ln 2\end{array}$

Now, $\begin{array}{rcl}\ln y& =& -\ln x+\frac{2}{x}+\ln (x-1)+c\end{array}$ will be

$\begin{array}{rcl}\ln y& =& -\ln x+\frac{2}{x}+\ln (x-1)+\ln 2\end{array}$

Step 3: Find the value of $y\left(4\right)$.

To find the value of $y\left(4\right)$, put $x=4$in the equation obtained from Step 2.

$\begin{array}{rcl}\ln y& =& -\ln 4+\frac{2}{4}+\ln (4-1)+\ln 2\\ & =& -\ln 4+\frac{1}{2}+\ln 3+\ln 2\\ & =& \ln \left(\frac{3}{2}\right)+\frac{1}{2}\ln e\end{array}$

Therefore, $y=\frac{3}{2}\left(\sqrt{e}\right)$.

Hence, option B is correct.