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

The Correct Option is A

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.