Question

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

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

Answer 1

Answer: (A)

Given differential equation is
$\left(y^2+2x\right)\frac{dy}{dx}=y<br/><br/>\Rightarrow \frac{dx}{dy}=\left(\frac{y^2+2x}{y}\right)$
$\Rightarrow \frac{dx}{dy}=y+\frac{2x}{y}$
$\Rightarrow \frac{dx}{dy}-\frac{2}{y}\cdot x=y$
$IF=e^{\int \frac{2}{y}dy}=e^{-2\log y}=y^{-2}=\frac{1}{y^2}$
- Complete solution is
$x\cdot \frac{1}{y^2}=\int y\cdot \frac{1}{y^2}dy+C$
$\Rightarrow \frac{x}{y^2}=\int \frac{dy}{y}+C={\log }_{e}y+C$
$\Rightarrow x=y^2{\log }_{e}y+c{y}^2$
$At+x=1,y=1$ then from Eq. (i), we get
$1=0+C\Rightarrow C=1$
$\therefore$ From Eq. (i), we get
$x=y^2\left({\log }_{e}y+1\right)$
Which is the required solution.