Question

If $x dy = y(dx+y dy),y(1)=1$ and $y(x) > 0$. Then, $y(-3)$ is equal to

• A. 3
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (A)

3

Answer 2

The correct option is(A): 3

Given, $x dy=y(dx+y dy), y > 0$
$\Rightarrow \, \, \, \, \, \, x \, dy-y \, dx=y^2dy$
$\Rightarrow \, \, \, \, \, \, \frac {x \, dy-y \, dx }{y^2}=dy \Rightarrow d\bigg ( \frac {x}{y} \bigg )=-dy$
On integrating both sides, we get
\hspace10mm \frac {x}{y}=-y+c \hspace10mm ...(i)
Since, $y(1)=1 \Rightarrow x=1,y=1$
\therefore \hspace8mm c=2
Now, E (i) becomes, $\frac {x}{y}+y = 2$
Again, for x=-3
\Rightarrow \hspace5mm -3+y^2=2y
\Rightarrow \hspace5mm y^2-2y-3=0
\Rightarrow \hspace5mm (y+1)(y-3)=0
As y>0, take y=3, neglecting y=-1.