Question

The solution of the differential equation $(y^2+2x)\frac{dy}{dx}=y$ satisfies $x=1,y=1$. Then the solution is:

• A. $x=y^2(1+{\log }_{e}y)$
• B. $y=x^2(1+{\log }_{e}x)$
• C. $x=y^2(1-{\log }_{e}y)$
• D. $y=x^2(1-{\log }_{e}x)$

Answer 1

Answer: (A)

$x=y^2(1+{\log }_{e}y)$

Given differential equation $(y^2+2x)\frac{dy}{dx}=y$ satisfies $x=1,y=1$
$\Rightarrow$ $\frac{dx}{dy}=y+\frac{2x}{y}$
$\Rightarrow$ $\frac{dx}{dy}-\frac{2}{y}.x=y$
If $e^{\int -\frac{2}{y}dy}=e^{-2\log y}=y^{-2}=\frac{1}{y^2}$
$\therefore$ Complete solution is

$x.\frac{1}{y^2}=\int y.\frac{1}{y^2}dy+C$

$\Rightarrow \frac{x}{y^2}=\int \frac{dy}{y}+C={\log }_{e}y+C$

$\Rightarrow x=y^2{\log }_{e}y+C{y}^2...\left(i\right)$

At $x=1$, $y=1$ then From eq. $\left(i\right)$ we get

$1=0+C$ $\Rightarrow$ $C=1$
$\therefore$ From eq $\left(i\right)$ we get
$x=y^2({\log }_{e}y+1)$