Question

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

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

Answer 1

Answer: (A)

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