Question

Solve the following differential equation :

$y\ln ydx+(x-\ln y)dy=0$

• A. $2x\times lny=l{n}^{2}y+c$
• B. $2y\times lny=l{n}^{2}x+c$
• C. $2x\times lny+l{n}^{2}y=c$
• D. $2y\times lny+l{n}^{2}y=c$

Answer 1

The correct option is A $2x\times lny=l{n}^{2}y+c$

$y\ln y+\frac{dy}{dx}(x-\ln y)=0$ ...(1)
Let $R(x,y)=y\ln y$ and $S(x,y)=x-\ln y$
This is not an exact equation, because $\frac{dR(x,y)}{dx}=\ln y+1\ne 1=\frac{dS(x,y)}{dx}$
Find an integrating factor $\mu$ such that
$\mu R(x,y)+\mu \frac{dy}{dx}S(x,y)=0$ is exact
This means $\frac{d}{dy}\left(\mu R(x,y)\right)=\frac{d}{dy}\left(\mu S(x,y)\right)$
$y\ln y\frac{d\mu }{dy}+\mu +\ln y\mu =\mu$
Isolate $\mu$ to the left hand side
$\frac{\frac{d\mu }{dy}}{\mu }=-\frac{1}{y}\Rightarrow \ln \mu =-\ln y\Rightarrow \mu =\frac{1}{y}$
Multiply both side of (1)
$\ln y+\frac{(x-\ln y)\frac{dy}{dx}}{y}=0$
Let $P(x,y)=\ln y$ and $Q(x,y)=\frac{x-\ln y}{y}$
This is exact equation, because $\frac{dP(x,y)}{dy}=\frac{1}{y}=\frac{dQ(x,y)}{dx}$
Define $f(x,y)$ such that $\frac{df(x,y)}{dx}=P(x,y)$ and $\frac{df(x,y)}{dy}=Q(x,y)$
Then, the solution will be given by $f(x,y)=c$
Integrate $\frac{df(x,y)}{dx}$ w.r.t x
$f(x,y)=\int \ln ydx=x\ln y+g\left(y\right)$ where $g\left(y\right)$ is arbitrary function of y
Differentiating $f(x,y)$ w.r.t y
$\frac{df(x,y)}{dy}=\frac{d}{dy}(x\ln y+g\left(y\right))=\frac{x}{y}+\frac{dg\left(y\right)}{dy}$
Substituting this in $\frac{df(x,y)}{dy}=Q(x,y)$
$\frac{x}{y}+\frac{dg\left(y\right)}{dy}=\frac{x-\ln y}{y}\Rightarrow \frac{dg\left(y\right)}{dy}=\frac{\ln y}{y}\Rightarrow g\left(y\right)=-\frac{1}{2}\ln {}^{2}y$
Hence
$-\frac{1}{2}\ln {}^{2}y+x\ln y=c$