Question

Solution of the diffferential equation $xdy=(y+x{y}^3(1+{\log }_{e}x\left)\right)dx$ is

• A. $\frac{-x^2}{y^2}=\frac{2}{3}{x}^3(\frac{2}{3}+{\log }_{e}x)+C$
• B. $\frac{x^2}{y^2}=\frac{2}{3}{x}^3(\frac{2}{3}+{\log }_{e}x)+C$
• C. $\frac{x^2}{y}=\frac{2}{3}{x}^3(\frac{2}{3}+{\log }_{e}x)+C$
• D. $\frac{-x^2}{y}=\frac{2}{3}{x}^3(\frac{2}{3}+{\log }_{e}x)+C$

Answer 1

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

$xdy=(y+x{y}^3(1+{\log }_{e}x\left)\right)dx\Rightarrow xdy-ydx=x{y}^3(1+{\log }_{e}x)dx\Rightarrow \frac{xdy-ydx}{y^2}=xy(1+{\log }_{e}x)dx\Rightarrow -d\left(\frac{x}{y}\right)=xy(1+{\log }_{e}x)dx\Rightarrow -\int \frac{x}{y}d\left(\frac{x}{y}\right)=\int x^2(1+{\log }_{e}x)dx+C\Rightarrow \frac{-x^2}{2y^2}=\frac{x^3}{3}(1+{\log }_{e}x)-\int \frac{x^2}{3}dx+C\Rightarrow \frac{-x^2}{2y^2}=\frac{x^3}{3}(1+{\log }_{e}x)-\frac{x^3}{9}+C\Rightarrow \frac{-x^2}{2y^2}=\frac{x^3}{3}(\frac{2}{3}+{\log }_{e}x)+C$

$\therefore \frac{-x^2}{y^2}=\frac{2}{3}{x}^3(\frac{2}{3}+{\log }_{e}x)+C$