Q.Show that the differential equation $(x - y)\frac{dy}{dx} = x + 2y$ is homogeneous and solve it.

Homogeneity: $\frac{dy}{dx}=\frac{x+2y}{x-y}=\frac{1+2(y/x)}{1-(y/x)}$ depends only on $y/x$, so the equation is homogeneous.

Put $y=vx$, $\frac{dy}{dx}=v+x\frac{dv}{dx}$:

Separate: $\dfrac{(1-v)\,dv}{v^2+v+1}=\dfrac{dx}{x}$. Split $1-v=-\tfrac12(2v+1)+\tfrac32$ and integrate:

Put $v=\frac{y}{x}$ (so $v^2+v+1=\frac{x^2+xy+y^2}{x^2}$); the $\log|x|$ terms cancel, and multiplying by $-2$:

Show it is homogeneous

Numerator and denominator are both degree 1, so dividing by $x$ leaves only the ratio $y/x$:

That is the homogeneous form.

Substitute $y=vx$

With $\frac{dy}{dx}=v+x\frac{dv}{dx}$,

Subtract $v$:

Separate

Integrate the left side

Split the numerator using the derivative of the denominator, $2v+1$:

Then

Completing the square, $v^2+v+1=\left(v+\tfrac12\right)^2+\tfrac34$, so

and the left integral is

Equating to $\int\frac{dx}{x}=\log|x|+C$:

Return to $x,y$

With $v=\frac{y}{x}$, $v^2+v+1=\dfrac{x^2+xy+y^2}{x^2}$, so

The $\log|x|$ terms cancel; multiplying by $-2$,

Method: Solving a homogeneous equation by $y = vx$

Use this when $\dfrac{dy}{dx} = \dfrac{M(x,y)}{N(x,y)}$ with $M$ and $N$ of the same degree, so the right side depends only on the ratio $\dfrac{y}{x}$.

Steps

Step 1: Confirm homogeneity.

Divide numerator and denominator by the appropriate power of $x$ until only $\dfrac{y}{x}$ appears; then $\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)$.

Step 2: Substitute $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.

Replace $y$ and $\dfrac{dy}{dx}$. The equation becomes an equation in $v$ and $x$.

Step 3: Subtract $v$ and separate.

Isolate $x\dfrac{dv}{dx}$, then split the variables into $\dfrac{(\text{function of }v)\,dv}{\cdots} = \dfrac{dx}{x}$.

Step 4: Integrate, then return to $x,y$.

For a rational integrand like $\dfrac{1-v}{v^2+v+1}$, write the numerator as a multiple of the denominator's derivative plus a constant, giving a $\log$ term plus an inverse-tangent (after completing the square). Finally substitute $v = \dfrac{y}{x}$ back.

Common Mistakes

Mistake 1: Forgetting to subtract $v$ after substituting.

Why it's wrong: $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$, so the plain $v$ must be moved to the right before separating; skipping it gives a wrong equation. Correct approach: isolate $x\dfrac{dv}{dx} = \dfrac{1+2v}{1-v} - v = \dfrac{v^2+v+1}{1-v}$.

Mistake 2: Integrating $\dfrac{1-v}{v^2+v+1}$ without splitting the numerator.

Why it's wrong: it is not a basic form; you must write $1-v = -\tfrac12(2v+1) + \tfrac32$ so one piece is a $\log$ and the other an inverse tangent (after completing the square). Correct approach: use the derivative of the denominator, $2v+1$, to split it.

Mistake 3: Not substituting $v = \dfrac{y}{x}$ back.

Why it's wrong: an answer left in $v$ is incomplete. Correct approach: replace $v$ with $\dfrac{y}{x}$ so the $\log|x|$ terms combine into $\log|x^2+xy+y^2|$.