Question

Solution of the equation $xdy-[y+x{y}^3(1+\log x)]dx=0$ is

• A. $\frac{-x^2}{y^2}=\frac{2x^3}{3}(\frac{2}{3}+\log x)+c$
• B. $\frac{x^2}{y^2}=\frac{2x^3}{3}(\frac{2}{3}+\log x)+c$
• C. $\frac{-x^2}{y^2}=\frac{x^3}{3}(\frac{2}{3}+\log x)+c$
• D. None of these

Answer 1

The correct option is A $\frac{-x^2}{y^2}=\frac{2x^3}{3}(\frac{2}{3}+\log x)+c$

We have, $xdy-ydx={xy}^3(1+\log x)dx$
$\Rightarrow -\left(\frac{ydx-xdy}{y^2}\right)=xy(1+\log x)dx$
$\Rightarrow -d\left(\frac{x}{y}\right)=xy(1+\log x)dx\Rightarrow -\frac{x}{y}d\left(\frac{x}{y}\right)=x^2(1+\log x)dx$
Integrating, we get
$\frac{{-\left(\frac{x}{y}\right)}^2}{2}=(1+\log x)\frac{x^3}{3}-\int \frac{x^3}{3}.\frac{1}{x}dx$
$\Rightarrow -\frac{x^2}{{2y}^2}=\frac{{2x}^3}{3}(1+\log x)\frac{x^3}{9}+\frac{c}{2}$
$\Rightarrow -\frac{x^2}{y^2}=\frac{{2x}^3}{3}(\frac{2}{3}+\log x)+c.$