Question

Solution of the differential equation $\frac{dy}{dx}=\frac{y}{2y\ln y+y-x},y>0$ is:
(where $C$ is integration constant)

• A. $x=y^2\ln y+\frac{C}{y^2}$
• B. $x=y\ln y+\frac{C}{y}$
• C. $x=y\ln y+C$
• D. $x=y^2\ln y+\frac{C}{y}$

Answer 1

The correct option is B $x=y\ln y+\frac{C}{y}$

The equation can be written as $\frac{dx}{dy}=\frac{2y\ln y+y-x}{y}=(2\ln y+1)-\frac{x}{y}$
$\Rightarrow \frac{dx}{dy}+\frac{1}{y}\cdot x=(2\ln y+1)$
In this equation it is clear that $P=\frac{1}{y}$ and $Q=(2\ln y+1).$
Which are both functions of $y$ only and equation contains derivatives of $x$ with repect to $y.$
I.F.$=e^{\int Pdy}=e^{\int \frac{1}{y}dy}=y$
$\therefore$ The solution is $x\left(\text{IF}\right)=\int (2\ln y+1)\left(\text{IF}\right)dy$
$\Rightarrow xy=\int (2\ln y+1)\cdot ydy$
On integrating By parts, and simplifying we get:
$xy=y^2\ln y+C$
$\Rightarrow x=y\ln y+\frac{C}{y}$