Q.Solve the differential equation $(x\,dy - y\,dx)\,y\sin\left(\frac{y}{x}\right) = (y\,dx + x\,dy)\,x\cos\left(\frac{y}{x}\right)$.

The key idea is that the equation is homogeneous — every term is of degree 2 in $x$ and $y$, so we substitute $y = vx$.

Step 1: Rewrite and substitute.

Divide both sides by $x^2$ (or rearrange directly). Let $y = vx$, so $dy = v\,dx + x\,dv$. The left side becomes:

Step 2: Simplify the right side.

Right side: $(vx\,dx + x(v\,dx + x\,dv))\, x \cos v = (2vx\,dx + x^2\,dv)\, x \cos v = (2v x^2\,dx + x^3\,dv) \cos v.$

Step 3: Equate and separate variables.

Bring terms with $dv$ together:

Factor $x^2$:

Cancel $x^2$ (for $x \neq 0$):

Step 4: Separate and integrate.

$\frac{v \sin v - \cos v}{v \cos v} \, dv = \frac{2}{x} \, dx.$ …

Why this approach works

When you see a differential equation where every term has the same total degree in $x$ and $y$, it's a homogeneous differential equation. The key insight: such equations can be simplified by writing $y = vx$, which turns the equation into one involving only $v$ and $x$. This works because the homogeneity lets us factor out powers of $x$ everywhere.

Look at our equation: every term has degree 2 in $x$ and $y$ (check: $x\,dy$ is degree 1+1=2, $y\,dx$ is degree 2, and the trigonometric functions are dimensionless). So the substitution $y = vx$ is the natural path.

Step-by-step solution

1. Rewrite the equation in a standard form

Start with:

Expand both sides:

2. Group terms with $dx$ and $dy$

Bring all terms to one side:

Factor $dy$ and $dx$:

3. Substitute $y = vx$ (so $dy = v\,dx + x\,dv$)

This is the heart of the method. Replace every $y$ with $vx$:

• $\frac{y}{x} = v$
• $\sin\left(\frac{y}{x}\right) = \sin v$, $\cos\left(\frac{y}{x}\right) = \cos v$
• $y^2 = v^2 x^2$

The equation becomes:

Simplify inside the brackets:

4. Divide through by $x^2$ (assuming $x \neq 0$)

5. Expand and collect $dx$ and $dv$ terms

Expand the first product:

Group $dx$ terms:

Simplify the $dx$ coefficient:

So we have:

6. Separate variables

Bring the $dv$ term to the other side:

Divide both sides by $x$ and by $v\cos v$ (careful: we'll handle special cases later):

7. Simplify the $dv$ side

So the separated equation is:

8. Integrate both sides

$\int \frac{2}{x}\,dx = \int \left(\tan v - \frac{1}{v}\right)dv$ …

Method: Homogeneous equation given in differential (M dx + N dy) form

Use this when the equation is presented as products of $dx$ and $dy$ (rather than as $\dfrac{dy}{dx} = \cdots$) but every term is the same degree in $x$ and $y$. Expand first, then use the homogeneous substitution.

Steps

Step 1: Expand all brackets and group the $dx$ and $dy$ terms.

Rewrite as $(\text{coeff})\,dx + (\text{coeff})\,dy = 0$ so the structure is visible.

Step 2: Substitute $y = vx$, with $dy = v\,dx + x\,dv$.

Replace $y$, $\dfrac{y}{x} = v$, and $dy$ everywhere. A common power of $x$ (here $x^2$) factors out and cancels.

Step 3: Collect and separate. …

Common Mistakes

Mistake 1: Substituting before expanding the brackets.

Why it's wrong: the equation is given as products with $dx$ and $dy$; putting $y=vx$ in without first expanding and grouping leads to tangled algebra. Correct approach: expand, collect $dx$ and $dy$ terms, then substitute.

Mistake 2: Handling $dy = v\,dx + x\,dv$ incorrectly.

Why it's wrong: forgetting the $x\,dv$ piece (or the common $x^2$ factor that cancels) derails the separation. Correct approach: substitute $dy$ fully and cancel the shared power of $x$. …