If $x \mathrm{~d} y=y(\mathrm{~d} x+y \mathrm{~d} y), y(1)=1, y(x)0$, then $y(-3)$ is

Question

If $x \mathrm{~d} y=y(\mathrm{~d} x+y \mathrm{~d} y), y(1)=1, y(x)>0$, then $y(-3)$ is

Aicharya · Accepted Answer

$x \mathrm{~d} y=y(\mathrm{~d} x+y \mathrm{~d} y)$ $\Rightarrow y \mathrm{~d} x=\left(x-y^2ight) \mathrm{d} y \Rightarrow \frac{\mathrm{d} x}{\mathrm{~d} y}+\left(-\frac{1}{y}ight) x=-y$ $	herefore \quad$ I.F. $=\mathrm{e}^{\int-\frac{1}{d y} y^y}=\mathrm{e}^{-\log y}=\frac{1}{y}$ $	herefore \quad$ Solution of the given equation is $x \cdot \frac{1}{y}=\int-y \cdot \frac{1}{y} \mathrm{~d} y+\mathrm{c}$ $\Rightarrow \frac{x}{y}=-y+\mathrm{c}$ ....(i) Since $y(1)=1$, i.e., $y=1$ when $x=1$ $	herefore \quad 1=-1+c \Rightarrow c=2$ $	herefore \quad \frac{x}{y}=-y+2 \quad \ldots[$ From (i) $]$ Putting $x=-3$, we get $-\frac{3}{y}=-y+2$ $\begin{aligned} & \Rightarrow y^2-2 y-3=0 \ & \Rightarrow(y-3)(y+1)=0\end{aligned}$ Since $y(x)>0, y=3$