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

Concept: Homogeneous Differential Equation — an equation of the form $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$.

Rewrite the given equation:

Divide through by $x\cos\left(\frac{y}{x}\right)$ (assuming $\cos(\frac{y}{x}) \neq 0$):

The right-hand side is a function of $\frac{y}{x}$ alone, so the equation is homogeneous.

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

Separate variables:

Integrate:

Replace $v = \frac{y}{x}$:

Why This Approach Works

A differential equation is homogeneous if every term in the numerator and denominator (when written as $\frac{dy}{dx} = \frac{M(x,y)}{N(x,y)}$) has the same total degree in $x$ and $y$. The key property: such equations can be transformed by the substitution $y = vx$, where $v = y/x$. This works because the function depends only on the ratio $y/x$, not on $x$ and $y$ separately. The substitution turns the equation into one where variables separate cleanly — you get $v$ on one side and $x$ on the other, and then integrate.

Let’s see this in action.

Step-by-Step Solution

1. Rewrite the equation in standard form.

We start with:

Divide both sides by $x\cos(y/x)$ (assuming $x \neq 0$ and $\cos(y/x) \neq 0$ for now):

Split the fraction:

Simplify:

So:

2. Confirm homogeneity.

A function $f(x,y)$ is homogeneous of degree $n$ if $f(tx, ty) = t^n f(x,y)$. Here, the right-hand side is $\frac{y}{x} + \sec(y/x)$. Replace $x$ with $tx$ and $y$ with $ty$:

No factor of $t$ appears — the function is homogeneous of degree 0. This confirms the substitution $y = vx$ will work.

3. Apply the substitution $y = vx$.

Let $v = y/x$, so $y = vx$. Differentiate with respect to $x$:

Substitute into the equation $\frac{dy}{dx} = v + \sec v$:

Cancel $v$ from both sides:

4. Separate variables.

Multiply both sides by $dx$ and divide by $x \sec v$ (or equivalently, multiply by $\cos v$):

Now the variables are separated — $v$ on the left, $x$ on the right.

5. Integrate both sides.

Integrate:

We get:

where $C$ is the constant of integration.

6. Substitute back $v = y/x$.

Replace $v$:

This is the general solution of the differential equation.

Method: Homogeneous equation, isolate $F(y/x)$ then substitute

Use this for equations where trigonometric or other functions of $\dfrac{y}{x}$ appear but every term still has the same degree, so the equation is homogeneous.

Steps

Step 1: Rearrange into $\dfrac{dy}{dx} = F\!\left(\dfrac{y}{x}\right)$.

Divide through by whatever multiplies $\dfrac{dy}{dx}$ and simplify until the right side is a function of the single ratio $\dfrac{y}{x}$. This confirms homogeneity of degree $0$.

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

Because the right side is exactly $F(v)$, the plain $v$ on the left usually cancels a matching $v$ on the right, leaving a clean $x\dfrac{dv}{dx} = (\text{function of }v)$.

Step 3: Separate and integrate.

Move the $v$-part to the left and $\dfrac{dx}{dv}$-part to the right, then integrate both standard pieces.

Step 4: Substitute $v = \dfrac{y}{x}$ to express the answer in the original variables.

Common Mistakes

Mistake 1: Not isolating $\dfrac{dy}{dx}$ before substituting.

Why it's wrong: you must first divide by $x\cos(y/x)$ to reach $\dfrac{dy}{dx} = \dfrac{y}{x} + \sec\!\left(\dfrac{y}{x}\right)$; substituting into the un-simplified form is messy and error-prone. Correct approach: get the clean $F(y/x)$ first.

Mistake 2: Missing the cancellation of $v$.

Why it's wrong: after $y=vx$ the equation becomes $v + x\dfrac{dv}{dx} = v + \sec v$, and the $v$ terms cancel to leave $x\dfrac{dv}{dx} = \sec v$. Forgetting this leaves an unsolvable-looking equation. Correct approach: cancel $v$, then separate as $\cos v\,dv = \dfrac{dx}{x}$.

Mistake 3: Leaving the answer in $v$.

Why it's wrong: the solution must be in $x$ and $y$. Correct approach: after integrating to $\sin v = \log|x| + C$, substitute $v = \dfrac{y}{x}$.