Question

If (1 + xy) $\frac{dy}{dx}+y^3=0$ and y(1) = 1, then the equation of the curve is

• A. $xy+y+1=2ye\frac{1}{y}-1$
• B. $xy+y+1=3ye\frac{1}{y}-1$
• C. $xy+y+1=2ye\frac{1}{y}+1$
• D. $xy+y+1=3ye\frac{1}{y}+1$

Answer 1

The correct option is B $xy+y+1=3ye\frac{1}{y}-1$

Given Equ. In $\frac{dx}{dy}+\frac{x}{y^2}=-\frac{1}{y^3}I.F.=e-\frac{1}{y}$
$\therefore x.e-\frac{1}{y}=\int -\frac{1}{y^3}.e-\frac{1}{y}dy$ and use y(1) = 1