Q.Solve the following differential equation: $y dx + x \log \left(\frac{y}{x}\right) dy - 2x dy = 0$

Both $M=y$ and $N=x\log(y/x)-2x$ are homogeneous of degree 1, so put $y=vx$, $dy=v\,dx+x\,dv$, $\log(y/x)=\log v$.

Divide by $x$ and collect $dx$:

With $t=\log v$ ($dt=\frac{dv}{v}$): $\int\frac{t-2}{t-1}\,dt=\int\left(1-\frac{1}{t-1}\right)dt=t-\log|t-1|$. So

$\log|x|+\log v-\log|\log v-1|=C.$ …

Why homogeneous

Write it as $M\,dx+N\,dy=0$ with

Every term is degree 1 in $x,y$, so the equation is homogeneous and $y=vx$ will separate it.

Substitute $y=vx$

With $dy=v\,dx+x\,dv$ and $\log(y/x)=\log v$,

Divide by $x$:

Collect the $dx$ terms: $v[1+\log v-2]=v(\log v-1)$, so

Separate

Divide by $x\,v(\log v-1)$:

Integrate

For the second term put $t=\log v$, $dt=\dfrac{dv}{v}$:

Hence

Return to $x,y$

With $v=\dfrac{y}{x}$, note $\log|x|+\log v=\log|x|+\log|y|-\log|x|=\log|y|$, so …

Method: Homogeneous equation better solved with $x=vy$

Use this when $y$ appears more simply than $x$ (e.g. logs of $y/x$ with $dy$ terms), as in $y\,dx+x\log\frac{y}{x}\,dy-2x\,dy=0$; substituting $x=vy$ can be cleaner than $y=vx$.

Steps

Step 1: Choose the tidier substitution

Because the equation is homogeneous, either $y=vx$ or $x=vy$ works; pick the one that simplifies the logarithm. With $x=vy$, $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$.

Step 2: Substitute and separate

Express everything in $v=\dfrac{x}{y}$, cancel matching terms, and separate the $v$ and $y$ parts. …

Common Mistakes

Mistake 1: Insisting on $y=vx$ when $x=vy$ is cleaner

Why it's wrong: with $\log\frac{y}{x}$ and $dy$ terms, $y=vx$ makes messy logs; $x=vy$ simplifies them. Correct approach: pick the substitution that tidies the ratio.

Mistake 2: Wrong derivative form for the chosen substitution …