Q.Solve the following differential equation: $x \frac{dy}{dx} - y + x \sin \left(\frac{y}{x}\right) = 0$

To determine if a differential equation $\frac{dy}{dx} = F(x, y)$ is homogeneous, we need to understand what a homogeneous function is.

A function $F(x, y)$ is called a homogeneous function of degree $n$ if, for any non-zero constant $\lambda$, the following condition holds:

For a differential equation $\frac{dy}{dx} = F(x, y)$ to be classified as a homogeneous differential equation, the function $F(x, y)$ must be a homogeneous function of degree zero. This means that when we replace $x$ with $\lambda x$ and $y$ with $\lambda y$, the function $F(x, y)$ must remain unchanged:

This property is crucial because it allows us to transform the differential equation into a separable form by substituting $y = vx$ (or $x = vy$). If $F(x, y)$ is homogeneous of degree zero, it can always be expressed as a function of $\frac{y}{x}$ (or $\frac{x}{y}$). For example, if $F(\lambda x, \lambda y) = F(x, y)$, we can choose $\lambda = \frac{1}{x}$ (assuming $x \neq 0$), then $F(x, y) = F\left(\frac{1}{x} \cdot x, \frac{1}{x} \cdot y\right) = F\left(1, \frac{y}{x}\right)$, which is clearly a function of $\frac{y}{x}$.

Let's examine each given option to see which $F(x, y)$ is not homogeneous of degree zero.

1. Option (A): $F(x, y) = \cos x - \sin\left(\dfrac{y}{x}\right)$ We test for homogeneity of degree zero by replacing $x$ with $\lambda x$ and $y$ with $\lambda y$:

For this to be equal to $F(x, y)$, we would need $\cos(\lambda x) = \cos x$. This is generally not true for arbitrary $\lambda \neq 1$. For instance, if $\lambda = 2$, then $\cos(2x) \neq \cos x$.
Therefore, $F(x, y) = \cos x - \sin\left(\dfrac{y}{x}\right)$ is **not** a homogeneous function of degree zero. This means the differential equation $\frac{dy}{dx} = \cos x - \sin\left(\dfrac{y}{x}\right)$ is not homogeneous.

2. Option (B): $F(x, y) = \dfrac{y}{x}$

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$:

This is equal to $F(x, y)$. Thus, $F(x, y) = \dfrac{y}{x}$ is a homogeneous function of degree zero.

3. Option (C): $F(x, y) = \dfrac{x^2 + y^2}{xy}$

Replace $x$ with $\lambda x$ and $y$ with $\lambda y$: …

Let’s understand why this equation fits the first order linear category, and why it does not fit the others.

A differential equation is called first order linear if it can be written in the form:

where $P(x)$ and $Q(x)$ are functions of $x$ only. The key idea is that $y$ and $\frac{dy}{dx}$ appear only to the first power, and there is no product like $y \cdot \frac{dy}{dx}$ or $y^2$.

Now, look at the given equation:

1. Isolate $\frac{dy}{dx}$ Divide the entire equation by $x \log x$ (provided $x > 0$ and $x \neq 1$, which is the natural domain for $\log x$):

2. Simplify the right-hand side Since $\frac{2 \log x}{x \log x} = \frac{2}{x}$ (cancelling $\log x$), we get:

3. Identify the form This is exactly $\frac{dy}{dx} + P(x) y = Q(x)$ with:

Both are functions of $x$ only, and $y$ appears linearly. So it is a first order linear differential equation.

Now, why are the other options wrong?

The idea is simple: a homogeneous differential equation is one where the right-hand side $F(x, y)$ doesn't change if you scale both $x$ and $y$ by the same factor. Why does that matter? Because if $F$ has that property, you can substitute $y = vx$ and turn the equation into one in $v$ and $x$ alone — a separable equation you can actually solve. The test is clean: check whether $F(tx, ty) = F(x, y)$ for any $t > 0$.

Let's go through each option.

1. Option (i): $F(x, y) = 3x + 2y$ Replace $x$ with $tx$ and $y$ with $ty$:

This is $t$ times the original, not equal to it — unless $t = 1$. So $F$ is homogeneous of degree 1, not degree 0. It also cannot be written as a function of $y/x$ alone (try it: $3x + 2y = x(3 + 2(y/x))$ still has an $x$ factor outside). So this is not homogeneous for the purpose of $\frac{dy}{dx} = F(x, y)$.

2. Option (ii): $F(x, y) = \sin\frac{y}{x} + \log y - \log x$ First simplify the log terms: $\log y - \log x = \log\frac{y}{x}$. So

This is already written purely in terms of $y/x$. That's a dead giveaway — it's homogeneous of degree 0. Check formally:

The $t$ cancels completely. So this is homogeneous.

3. Option (iii): $F(x, y) = e^{y/x} + 1$ Again, this is already a function of $y/x$ alone. …