Question

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

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $3$

$xdy=ydx+y^{2}dy$

$-y^{2}dy=ydx+y^{2}dy$

$dy=\frac{ydx-xdy}{-y^2}$ (usiny formula $\frac{ydx-xdy}{-y^2}=d\left(\frac{x}{y}\right)$)

$dy=-\left(d\left(\frac{x}{y}\right)\right)$

Integrating both side worot x

$\begin{array}{c}\Rightarrow y=\frac{-x}{y}+c_1\\ now,\\ y\left(I\right)=1Ifx=1,y=1\\ \Rightarrow I=\frac{-1}{I}+c_1{c}_1=2\\ y=\frac{-x}{y}+2\\ Ifx=-3\\ \Rightarrow y=\frac{+3}{y}+2\\ \Rightarrow y^2-2y-3=0\\ \Rightarrow y^2-3y+y-3=0\\ \Rightarrow y\left(y-3\right)+1\left(y-3\right)=0\\ \Rightarrow \left(y-3\right)\left(y+1\right)=0\\ y=3,-1\end{array}$

Hence,

$y=3$

Option $A$ is correct answer.