Question

Find the general solution of the differential equation \(y - x \frac{dy}{dx} = x + y \frac{dy}{dx}\).

Hint:

A differential equation of the form \(\frac{dy}{dx} = F\left(\frac{y}{x}\right)\) is homogeneous. The standard substitution \(y = vx\) will always transform it into a separable equation in terms of \(v\) and \(x\). Always remember to substitute back to the original variables \(x\) and \(y\) after integration.

Answer 1

Step 1: Understanding the Concept:
The given equation is a first-order ordinary differential equation. We need to identify its type to choose the correct method of solution. By rearranging the equation, we can determine if it is separable, homogeneous, linear, or exact.
Step 2: Key Formula or Approach:
First, we rearrange the equation to isolate the derivative term \(\frac{dy}{dx}\).
\[ y - x = x \frac{dy}{dx} + y \frac{dy}{dx} \] \[ y - x = (x + y) \frac{dy}{dx} \] \[ \frac{dy}{dx} = \frac{y - x}{y + x} \] This is a homogeneous differential equation because the numerator and denominator are homogeneous functions of the same degree (degree 1). To solve it, we use the substitution \(y = vx\), which implies \(\frac{dy}{dx} = v + x \frac{dv}{dx}\).
Step 3: Detailed Explanation:
Substitute \(y = vx\) and \(\frac{dy}{dx} = v + x \frac{dv}{dx}\) into the equation:
\[ v + x \frac{dv}{dx} = \frac{vx - x}{vx + x} \] Factor out \(x\) from the numerator and denominator on the right side:
\[ v + x \frac{dv}{dx} = \frac{x(v - 1)}{x(v + 1)} = \frac{v - 1}{v + 1} \] Now, we separate the variables \(v\) and \(x\).
\[ x \frac{dv}{dx} = \frac{v - 1}{v + 1} - v \] \[ x \frac{dv}{dx} = \frac{v - 1 - v(v + 1)}{v + 1} = \frac{v - 1 - v^2 - v}{v + 1} \] \[ x \frac{dv}{dx} = \frac{-v^2 - 1}{v + 1} = -\frac{v^2 + 1}{v + 1} \] Rearrange to separate the variables:
\[ \frac{v + 1}{v^2 + 1} dv = -\frac{1}{x} dx \] Integrate both sides:
\[ \int \frac{v + 1}{v^2 + 1} dv = -\int \frac{1}{x} dx \] Split the integral on the left side:
\[ \int \frac{v}{v^2 + 1} dv + \int \frac{1}{v^2 + 1} dv = -\int \frac{1}{x} dx \] Evaluate each integral:
The first integral is \(\frac{1}{2}\ln(v^2 + 1)\).
The second integral is \(\tan^{-1}(v)\).
The integral on the right is \(-\ln|x|\).
So, the solution in terms of \(v\) and \(x\) is:
\[ \frac{1}{2}\ln(v^2 + 1) + \tan^{-1}(v) = -\ln|x| + C \] where C is the constant of integration.
Now, substitute back \(v = \frac{y}{x}\):
\[ \frac{1}{2}\ln\left(\left(\frac{y}{x}\right)^2 + 1\right) + \tan^{-1}\left(\frac{y}{x}\right) = -\ln|x| + C \] Simplify the logarithm term:
\[ \frac{1}{2}\ln\left(\frac{y^2 + x^2}{x^2}\right) + \tan^{-1}\left(\frac{y}{x}\right) = -\ln|x| + C \] \[ \frac{1}{2}[\ln(y^2 + x^2) - \ln(x^2)] + \tan^{-1}\left(\frac{y}{x}\right) = -\ln|x| + C \] \[ \frac{1}{2}\ln(y^2 + x^2) - \frac{1}{2}(2\ln|x|) + \tan^{-1}\left(\frac{y}{x}\right) = -\ln|x| + C \] \[ \frac{1}{2}\ln(y^2 + x^2) - \ln|x| + \tan^{-1}\left(\frac{y}{x}\right) = -\ln|x| + C \] The \(-\ln|x|\) terms on both sides cancel out.
\[ \frac{1}{2}\ln(y^2 + x^2) + \tan^{-1}\left(\frac{y}{x}\right) = C \] Step 4: Final Answer:
The general solution of the differential equation is \(\frac{1}{2}\ln(y^2 + x^2) + \tan^{-1}\left(\frac{y}{x}\right) = C\).