Question

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx}=\frac{y}{x}\left(1+x{y}^2\left(1+{\log }_{e}x\right)\right),x>0,y(1)=3$. Then $\frac{y^2(x)}{9}$ is equal to :

• A. $\frac{x^2}{5-2x^3\left(2+{\log }_e{x}^3\right)}$
• B. $\frac{x^2}{3x^3\left(1+{\log }_e{x}^2\right)-2}$
• C. $\frac{x^2}{7-3x^3\left(2+{\log }_e{x}^2\right)}$
• D. $\frac{x^2}{2x^3\left(2+{\log }_e{x}^3\right)-3}$

Answer 1

Answer: (A)

$\frac{dy}{dx}-\frac{y}{x}=y^3\left(1+{\log }_{e}x\right)$
$\frac{1}{y^3}\frac{dy}{dx}-\frac{1}{x{y}^2}=1+{\log }_{e}x$
$\text{ Let }-\frac{1}{y^2}=t\Rightarrow \frac{2}{y^3}\frac{dy}{dx}=\frac{dt}{dx}$
$\therefore \frac{dt}{dx}+\frac{2t}{x}=2\left(1+{\log }_{e}x\right)$
$\text{ I.F. }=e^{\int \frac{2}{x}dx}=x^2$
$\frac{-x^2}{y^2}=\frac{2}{3}\left(\left(1+{\log }_{e}x\right)x^3-\frac{x^3}{3}\right)+C$
$y(1)=3$
$\frac{y^2}{9}=\frac{x^2}{5-2x^3\left(2+{\log }_e{x}^3\right)}$
OR
$xdy=ydx+x^3\left(1+{\log }_{e}x\right)dx$
$\frac{xdy-ydx}{y^3}=x\left(1+{\log }_{e}x\right)dx$
$-\frac{x}{y}d\left(\frac{x}{y}\right)=x^2\left(1+{\log }_{e}x\right)dx$
$-{\left(\frac{x}{y}\right)}^2=2\int x^2\left(1+{\log }_{e}x\right)dx$