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

The equation is homogeneous. Writing it as $\dfrac{dy}{dx}$ and putting $y=vx$ (so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$):

Then

Separating variables gives $\left(\tan v-\dfrac{1}{v}\right)dv=\dfrac{2\,dx}{x}$. Integrating: …

Idea

Every term contains $y/x$, and both sides are homogeneous of the same degree, so the ratio $y/x$ is the natural variable. Substituting $y=vx$ collapses the messy trig factors into something separable.

Set up

First write the equation in derivative form. Dividing the $y\,dx$ side by the $x\,dy$ side,

Put $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ and $\dfrac{y}{x}=v$. Substituting and cancelling $x^{2}$ from top and bottom:

Work the steps

1. Isolate $x\dfrac{dv}{dx}$:

2. Separate the variables:

3. Integrate both sides. Using $\int \tan v\,dv=-\log|\cos v|$ and $\int \dfrac{dv}{v}=\log|v|$: …

Method: Homogeneous equation with trig functions of $y/x$

Use this for equations whose coefficients are built from $\cos\frac{y}{x}$ and $\sin\frac{y}{x}$; they are homogeneous of degree 1, so $y=vx$ makes them separable.

Steps

Step 1: Solve for $\dfrac{dy}{dx}$ and substitute $y=vx$

Write the equation as $\dfrac{dy}{dx}=F(v)$ with $v=\dfrac{y}{x}$, then use $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.

Step 2: Separate and simplify the trig integrand

After cancelling, the $v$-side collects $\sin,\cos$ terms; using identities and recognising a $\dfrac{d}{dv}$ pattern makes it integrable. …

Common Mistakes

Mistake 1: Not checking that the equation is homogeneous of degree 1

Why it's wrong: only then does $y=vx$ apply; the trig factors $\sin\frac{y}{x},\cos\frac{y}{x}$ are degree-0, so each bracket times $x$ or $y$ is degree 1. Correct approach: confirm the degree before substituting.

Mistake 2: Mishandling the trig algebra after substitution …