Question Statement

Solve the differential equation:

$(x-y)dx+xdy=0$

---

Background and Explanation

This is a homogeneous differential equation, where the substitution $y=ux$ (or $y=vx$) transforms it into a separable equation. A function is homogeneous of degree zero if it can be expressed as a function of the ratio $\frac{y}{x}$.

---

Solution

First, we rearrange the given equation to express $\frac{dy}{dx}$:

\begin{align*} (x-y) \, dx + x \, dy &= 0 \\ x \, dy &= -(x-y) \, dx \\ \frac{dy}{dx} &= \frac{-(x-y)}{x} \\ \frac{dy}{dx} &= \frac{-x + y}{x} \end{align*}

Since the right-hand side can be written as $\frac{y}{x}-1$, this confirms the equation is homogeneous (function of $\frac{y}{x}$).

Substitution: Let $y=ux$, where $u$ is a function of $x$.

Differentiating with respect to $x$ using the product rule:

$\frac{dy}{dx}=x\frac{du}{dx}+u$

Substituting into equation (1):

\begin{align*} x\frac{du}{dx} + u &= \frac{-x + ux}{x} \\ x\frac{du}{dx} + u &= \frac{x(-1 + u)}{x} \\ x\frac{du}{dx} + u &= -1 + u \end{align*}

Subtracting $u$ from both sides:

\begin{align*} x\frac{du}{dx} &= -1 \\ x \, du &= -dx \\ du &= -\frac{dx}{x} \end{align*}

Integrating both sides:

\begin{align*} \int du &= \int -\frac{1}{x} \, dx \\ u &= -\ln|x| + C \end{align*}

Rearranging and substituting back $u=\frac{y}{x}$:

\begin{align*} u + \ln|x| &= C \\ \frac{y}{x} + \ln|x| &= C \end{align*}

Multiplying through by $x$ gives the general solution:

$y=x(C-\ln |x|)$

Or equivalently: $\ln |x|+\frac{y}{x}=C$

---

Key Formulas or Methods Used

• Homogeneous equation test: $\frac{dy}{dx}=F\left(\frac{y}{x}\right)$
• Substitution for homogeneous equations: $y=ux⟹\frac{dy}{dx}=x\frac{du}{dx}+u$
• Separation of variables: $\frac{dy}{dx}=f(x)g(y)⟹\int \frac{dy}{g(y)}=\int f(x)dx$
• Basic integration: $\int \frac{1}{x}dx=\ln |x|+C$

---

Summary of Steps

1. Rearrange the equation to isolate $\frac{dy}{dx}=\frac{y-x}{x}$ and verify it's homogeneous
2. Substitute $y=ux$ and differentiate to get $\frac{dy}{dx}=x\frac{du}{dx}+u$
3. Simplify the resulting equation to obtain $x\frac{du}{dx}=-1$
4. Separate variables: $du=-\frac{dx}{x}$
5. Integrate both sides to get $u=-\ln |x|+C$
6. Back-substitute $u=\frac{y}{x}$ to obtain the final solution $\frac{y}{x}+\ln |x|=C$