Question

The solution of $(1+y+x^{2}y)dx+(x+x^3)dy=0$ is:

• A. $y+{\tan }^{-1}x=C$
• B. $xy+{\tan }^{-1}x=C$
• C. $y^2+{\tan }^{-1}x=C$
• D. $x^2+{\tan }^{-1}y/x=C$

Answer 1

The correct option is B $xy+{\tan }^{-1}x=C$

$(1+y+x^{2}y)dx+(x+x^3)dy=0$
$\Rightarrow \frac{dy}{dx}+\frac{(x^2+1)y}{x^3+x}=-\frac{1}{x^3+x}$
Let $\mu =e^{\int \frac{(x^2+1)}{x^3+x}dy}=x$
Multiplying both sides by $\mu$
$\frac{xdy}{dx}+\frac{x(x^2+1)y}{x^3+x}=-\frac{1}{x^3+x}$
$\Rightarrow \frac{xdy}{dx}+\frac{d}{dx}\left(x\right)y=-\frac{x}{x^3+x}$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dx}\left(xy\right)=-\frac{x}{x^3+x}$
Integrating both sides
$\Rightarrow \int \frac{d}{dx}\left(xy\right)dx=-\int \frac{x}{x^3+x}dx\Rightarrow xy=-{\tan }^{-1}x=c$