Let $y = y(x)$ be the solution of the differential equation ${x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}$. Then y (1) is equal to

$x^{3} d y+x y d x-d x=0$ <br/><br/> $\Rightarrow \frac{d y}{d x}=\frac{1-x y}{x^{3}}$ <br/><br/> $\Rightarrow \frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}$ <br/><br/> I.F. $=e^{\int \frac{d x}{x^{2}}}=e^{-\frac{1}{x}}$ <br/><br/> $\therefore \quad y e^{-\frac{1}{x}}=\int \frac{e^{-\frac{1}{x}}}{x^{3}} d x$ <br/><br/> For RHS put $-\frac{1}{x}=t \Rightarrow \frac{d x}{x^{2}}=d t$ <br/><br/> $\therefore y e^{-\frac{1}{x}}=-\int t e^{t} d t$ <br/><br/> $\Rightarrow y e^{-\frac{1}{x}}=-\left[t e^{t}-e^{t}\right]+c$ <br/><br/> $\Rightarrow y e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x}+e^{-\frac{1}{x}}+c$ <br/><br/> $\downarrow y\left(\frac{1}{2}\right)=3-e$ <br/><br/> $\Rightarrow(3-e) e^{-2}=2 e^{-2}+e^{-2}+c$ <br/><br/> $\Rightarrow \quad c=-\frac{1}{e}$ <br/><br/> For $y(1)$ put $x=1, c=-e^{-1}$ in equation (i) we get $y e^{-1}=e^{-1}+e^{-1}-e^{-1}$ <br/><br/> $\Rightarrow y=1$