Question

The integrating factor of the differential equation $(2x + 3y^2) dy = y \,dx (y > 0) $ is

• A. $\frac{1}{x} $
• B. $\frac{1}{_ey} $
• C. $\frac{1}{y^2} $
• D. $ - \frac{1}{y^2} $

Answer 1

Answer: (C)

$\left(2x+3y^{2}\right)dy=y dx$
$\Rightarrow \, \frac{2x+3y^{2}}{y}=\frac{d x}{d y} \quad\Rightarrow \, \frac{2x}{y}+3y=\frac{d x}{d y} \quad\Rightarrow \, \frac{d x}{d y}-\frac{2}{y}x=3y$
$If==e^{\int-\frac{2}{y}dy} =e^{-2 log y} =e^{log y^{-2}}=\frac{1}{y^{2}}$