Q94 / 200 · Manipal (MET) 2012 · Mathematics
If $x d y=y(d x+y d y), y(1)=1$ and $y(x)\gt0$. Then, $y(-3)$ is equal to
- Correct answerA3
- B2
- C1
- D0
Solution
$\begin{array}{ll}x d y=y(d x+y d y), y\gt0 \\ \Rightarrow & x d y-y d x=y^2 d y \\ \Rightarrow & \frac{x d y-y d x}{y^2}=d y\end{array}$
$\Rightarrow \quad-d\left(\frac{x}{y}\right)=d y$
On integrating both sides, we get
$\frac{x}{y}=-y+C \qquad \text{....(i)}$
As $\quad y(1)=1 \Rightarrow x=1, y=1$
$\therefore \quad C=2$
$\therefore$ Eq. (i) becomes
$\frac{x}{y}+y=2$
Again, for $\quad x=-3$
$\begin{array}{lr}
\Rightarrow & -3+y^2=2 y \\
\Rightarrow & y^2-2 y-3=0 \\
\Rightarrow & (y+1)(y-3)=0
\end{array}$
As $y\gt0$, we take $y=3$, (neglecting $y=-1$ )
Hence, the correct answer is option: 'A'.