Q.Solve the following differential equation: $(x - y) dy - (x + y) dx = 0$

The key idea is that this is a homogeneous differential equation — the coefficients of $dx$ and $dy$ are homogeneous functions of the same degree (degree 1).

Step 1: Rewrite in standard form

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

Step 3: Separate variables

Step 4: Integrate both sides

Replace $v = y/x$:

Simplify the log term:

Cancel $\log|x|$ on both sides:

Why the homogeneous approach works

A differential equation of the form $M(x,y)\,dx + N(x,y)\,dy = 0$ is homogeneous if $M$ and $N$ are homogeneous functions of the same degree — meaning each term has the same total power of $x$ and $y$. Here, every term in $(x-y)dy - (x+y)dx = 0$ is of degree 1: $x$, $y$ each appear linearly.

The key insight: when both $M$ and $N$ are homogeneous of degree $n$, the ratio $\frac{dy}{dx}$ depends only on $\frac{y}{x}$. That lets us substitute $y = vx$, turning the equation into one where variables separate cleanly. No guesswork — it’s a standard, reliable method.

Step-by-step solution

1. Rewrite in standard form

Start with

Bring the $dx$ term to the other side:

Divide through by $dx$ (assuming $x \neq 0$):

So

This is our working form.

2. Confirm homogeneity

The right-hand side is a ratio of two degree‑1 expressions. Divide numerator and denominator by $x$:

It depends only on $v = \frac{y}{x}$ — homogeneous, confirmed.

3. Substitute $y = vx$

Let $y = vx$, where $v$ is a function of $x$. Then

Plug into the equation:

4. Separate variables

Subtract $v$ from both sides:

Combine the right-hand side over a common denominator:

Thus

Separate:

5. Integrate both sides

Left side:

The first integral is $\tan^{-1} v$. For the second, let $u = 1 + v^2$, so $du = 2v\,dv$ and $v\,dv = \frac{du}{2}$:

(Since $1+v^2 > 0$, absolute value is unnecessary.)

Right side:

Putting it together:

6. Back-substitute $v = \frac{y}{x}$

Simplify the log term:

So

Substitute back:

The $-\log|x|$ on the left and $+\log|x|$ on the right cancel:

7. Rearrange for the final form

(Note: the constant $C$ is arbitrary; its sign is absorbed.)

Method: Homogeneous equation giving an $\arctan$–log answer

Use this for homogeneous equations such as $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}$, where after $y=vx$ the $v$-integral splits into an inverse-tangent part and a logarithm part.

Steps

Step 1: Substitute $y=vx$ and simplify

Replace to get $v+x\dfrac{dv}{dx}=\dfrac{1+v}{1-v}$, then isolate $x\dfrac{dv}{dx}=\dfrac{1+v^2}{1-v}$.

Step 2: Separate and split the numerator

$\dfrac{1-v}{1+v^2}\,dv=\dfrac{dx}{x}$; split the left side into $\dfrac{1}{1+v^2}-\dfrac{v}{1+v^2}$.

Step 3: Integrate and back-substitute

These give $\tan^{-1}v-\tfrac12\log(1+v^2)=\log|x|+C$; put $v=\dfrac{y}{x}$ to reach $\tan^{-1}\dfrac{y}{x}=\tfrac12\log(x^2+y^2)+C$.

Common Mistakes

Mistake 1: Not splitting $\dfrac{1-v}{1+v^2}$ into two integrals

Why it's wrong: it must break into $\frac{1}{1+v^2}$ (gives $\tan^{-1}v$) and $\frac{v}{1+v^2}$ (gives $\frac12\log(1+v^2)$); treating it as one blocks integration. Correct approach: separate the two pieces.

Mistake 2: Sign error on the logarithm term

Why it's wrong: the middle sign is minus, so you get $\tan^{-1}v-\frac12\log(1+v^2)$; a sign slip changes the final relation. Correct approach: carry the minus through.

Mistake 3: Not converting $1+v^2$ back to $\dfrac{x^2+y^2}{x^2}$

Why it's wrong: skipping this leaves the answer in $v$ instead of the clean $\tan^{-1}\frac{y}{x}=\frac12\log(x^2+y^2)+C$. Correct approach: back-substitute $v=\frac{y}{x}$ and simplify the log.