# xydx+\{\frac{1}{2}x^{2}(1-y \ln y-y)\} dy=0

$xydx+\left\{\frac{1}{2}{x}^{2}\left(1-y\mathrm{ln}y-y\right)\right\}dy=0$
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Gerald Lopez
Step 1
The differential equation is $xydx+\left\{\frac{1}{2}{x}^{2}\left(1-y\mathrm{ln}y-y\right)\right\}dy=0$
Rewrite the equation in the form of separable equation as follows.
$xydx+\left\{\frac{1}{2}{x}^{2}\left(1-y\mathrm{ln}y-y\right)\right\}dy=0$
$xy+\left\{\frac{1}{2}{x}^{2}\left(1-y\mathrm{ln}y-y\right)\right\}\frac{dy}{dx}=0$ (Divide by dx on both sides)
$\frac{1}{2}{x}^{2}\left(1-y\mathrm{ln}y-y\right)\frac{dy}{dx}=-xy$
$\frac{\left(1-y\mathrm{ln}y-y\right)}{y}\frac{dy}{dx}=\frac{-2x}{{x}^{2}}$
$\frac{\left(1-y\mathrm{ln}y-y\right)}{y}\frac{dy}{dx}=-\frac{2}{x}$
$\frac{\left(1-y\mathrm{ln}y-y\right)}{y}dy=-\frac{2}{x}dx$
Step 2
Now, integrate both sides of the obtained equation and find the solution of the differential equation.
$\int \frac{\left(1-y\mathrm{ln}y-y\right)}{y}dy=\int -\frac{2}{x}dx$
$\int \frac{1}{y}dy-\int \mathrm{ln}\left(y\right)dy-\int dy=-2\int \frac{1}{x}dx$
$\mathrm{ln}\left(y\right)-y\mathrm{ln}\left(y\right)+y-y=-2\mathrm{ln}\left(x\right)+C$
$\mathrm{ln}\left(y\right)-y\mathrm{ln}\left(y\right)+2\mathrm{ln}\left(x\right)=C$
$-y\mathrm{ln}\left(y\right)+\mathrm{ln}\left(y\right)+\mathrm{ln}\left({x}^{2}\right)=C$
Donald Cheek
Step 1
The given differntial equation is $xydx+\left(\frac{1}{2}{x}^{2}\left(1-y\mathrm{ln}y-y\right)\right)dy=0$.
It can be seen that it is the first order seperable differntial equation.
Now seperate the variables.
$\left(-\mathrm{ln}\left(y\right)-1+\frac{1}{y}\right)dy=-\frac{2}{x}dx$
Step 2
Integrate both sides.
$\int \left(-\mathrm{ln}\left(y\right)-1+\frac{1}{y}\right)dy=\int -\frac{2}{x}dx$
$\int \left(-\mathrm{ln}\left(y\right)\right)dy+\int \left(-1+\frac{1}{y}\right)dy=\int -\frac{2}{x}dx$
$-\left(y\mathrm{ln}\left(y\right)-\int 1dy\right)+\int \left(-1\right)dy+\int \frac{1}{y}dy=-\int \frac{2}{x}dx$ using by part $\left\{u=\mathrm{ln}y,{v}^{\prime }=1\right\}$
$-y\mathrm{ln}y+\int \frac{1}{y}dy=-\int \frac{2}{x}dx$
$-y\mathrm{ln}y+\mathrm{ln}y=-2\mathrm{ln}x+{c}_{1}$
$-y\mathrm{ln}y+\mathrm{ln}y2\mathrm{ln}x={c}_{1}$

Answer: The solution of the diffrential equation is $-\mathrm{ln}\left(y\right)y+\mathrm{ln}\left({x}^{2}y\right)={c}_{1}$
###### Not exactly what you’re looking for?
user_27qwe

Given: $xydx+\left\{\frac{1}{2{x}^{2}}\left(1-y\mathrm{ln}y-y\right)\right\}dy=0$
The eqn can be written as
$xydx=-\frac{1}{2{x}^{2}}\left(1-y\mathrm{ln}y-y\right)dy$
or $2{x}^{3}dx=-\left(\frac{1}{y}-\mathrm{ln}y-1\right)dy$
$2{x}^{3}dx=\left(1+\mathrm{ln}y-\frac{1}{y}\right)dy$
Integrating both sides
$2\int {x}^{3}dx=\int \left(1+\mathrm{ln}y-\frac{1}{y}\right)dy+c$
$\frac{{x}^{4}}{2}=y+\int \mathrm{ln}ydy-\mathrm{ln}y+c$ (1)
Here $\int \mathrm{ln}y=\int 1\cdot \mathrm{ln}y=y\mathrm{ln}y-\int \frac{1}{y}\cdot ydy$
$=y\mathrm{ln}y-y\left(1\right)$
$\frac{{x}^{2}}{2}=y+y\mathrm{ln}y-y-\mathrm{ln}y+c$
or $\frac{{x}^{2}}{2}=\left(y-1\right)\mathrm{ln}y+c$