Question Statement

Solve the differential equation:

$\left[y+x\cot \left(\frac{y}{x}\right)\right]dx-xdy=0$

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Background and Explanation

This is a homogeneous differential equation, meaning $\frac{dy}{dx}$ can be expressed purely as a function of the ratio $\frac{y}{x}$. Such equations are solved using the substitution $y=ux$, which transforms them into separable equations through the product rule.

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Solution

Start with the given differential equation:

$\left[y+x\cot \left(\frac{y}{x}\right)\right]dx-xdy=0$

Rearrange to isolate the differential terms:

$\left[y+x\cot \left(\frac{y}{x}\right)\right]dx=xdy$

Divide both sides by $x$ to obtain the derivative form $\frac{dy}{dx}$:

\begin{align*} \frac{y}{x} + \frac{x \cot \left(\frac{y}{x}\right)}{x} &= \frac{d y}{d x} \\ \frac{d y}{d x} &= \frac{y}{x} + \cot \left(\frac{y}{x}\right) \end{align*}

Equation (1) confirms this is a homogeneous differential equation. Apply the standard substitution:

$u=\frac{y}{x}\Rightarrow y=ux$

Differentiate $y=ux$ with respect to $x$ using the product rule:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{du}{dx}\cdot x+u\cdot 1\\ \frac{dy}{dx}& =x\frac{du}{dx}+u\end{array}$

Substitute this expression into equation (1):

$\begin{array}{rl}x\frac{du}{dx}+u& =u+\cot u\\ x\frac{du}{dx}& =u+\cot u-u\\ x\frac{du}{dx}& =\cot u\end{array}$

Separate the variables by moving all $u$ terms to the left and $x$ terms to the right. Recall that $\frac{1}{\cot u}=\tan u$:

$\begin{array}{rl}\frac{du}{\cot u}& =\frac{dx}{x}\\ \tan udu& =\frac{dx}{x}\end{array}$

Integrate both sides:

\begin{align*} \int \tan u \, d u &= \int \frac{d x}{x} \\ \ln (\sec u) &= \ln x + \ln c \\ \ln (\sec u) &= \ln (c x) \\ \sec u &= c x \end{align*}

Substitute back $u=\frac{y}{x}$ into equation (2) to obtain the general solution:

$\sec \left(\frac{y}{x}\right)=cx$

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Key Formulas or Methods Used

• Homogeneous substitution: $u=\frac{y}{x}$ or equivalently $y=ux$
• Product rule for differentiation: $\frac{d}{dx}(ux)=u+x\frac{du}{dx}$
• Trigonometric identity: $\frac{1}{\cot u}=\tan u$
• Integration of tangent: $\int \tan udu=\ln |\sec u|+C$ (or $-\ln |\cos u|+C$)
• Logarithm properties: $\ln a+\ln b=\ln (ab)$ and $e^{\ln x}=x$
• Separation of variables technique

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Summary of Steps

1. Rearrange the equation to express $\frac{dy}{dx}$ explicitly on one side
2. Identify the equation as homogeneous (right-hand side is a function of $\frac{y}{x}$ only)
3. Substitute $u=\frac{y}{x}$ and apply $\frac{dy}{dx}=u+x\frac{du}{dx}$
4. Simplify the resulting equation by canceling terms to isolate $x\frac{du}{dx}$
5. Separate variables: $\tan udu=\frac{dx}{x}$
6. Integrate both sides to obtain $\ln |\sec u|=\ln |cx|$
7. Exponentiate to remove logarithms, then substitute back $u=\frac{y}{x}$ for the final solution $\sec \left(\frac{y}{x}\right)=cx$