Question

If $xdy=y(dx+ydy)$, $y\left(1\right)=1$ and $y\left(x\right)>0$, then $y(-3)$ is equal to?

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A

$3$

Explanation for the correct option.

Step 1: Solve the given equation $xdy=y(dx+ydy)$.

$$\begin{gathered} xdy=y(dx+ydy) \\ \Rightarrow xdy=ydx+y^{2}dy \\ \Rightarrow xdy-ydx=y^{2}dy \\ \Rightarrow \frac{xdy-ydx}{y^2}=dy \\ \Rightarrow d\left(\frac{x}{y}\right)=-dy\left[by\frac{ydx-xdy}{-y^2}=d\left(\frac{x}{y}\right)\right] \end{gathered}$$

Integrating both sides with respect to $x$, we get

$\frac{x}{y}=-y+c.....\left(1\right)$

Step 2: Find the value of $c$.

By substituting the value of $y\left(1\right)$in $\left(1\right)$, we get

$\begin{array}{rcl}c& =& \frac{x}{y}+y\\ & =& \frac{1}{1}+1\\ & =& 2\end{array}$

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

$$\begin{gathered} \frac{x}{y}=-y+c \\ \Rightarrow \frac{-3}{y}=-y+2 \\ \Rightarrow y^2-2y-3=0 \\ \Rightarrow y^2+y-3y-3=0 \\ \Rightarrow \left(y-3\right)\left(y+1\right)=0 \end{gathered}$$

Now, $y=3\text{or}-1$. As $y\left(x\right)>0$,

So, $y=3$

Hence, option (A) is correct.