Question:
Find the general solution of the differential equation \( (x^2 + y^2) \, dy = xy \, dx \).
Find the general solution of the differential equation \( (x^2 + y^2) \, dy = xy \, dx \).
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In homogeneous equations, look out for log properties during simplification. Combining \( \log|v| + \log|x| = \log|vx| \) simplifies back to \( \log|y| \) directly, saving major back-substitution algebra steps.
Updated On: Jul 4, 2026
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Solution and Explanation
Concept:
The given differential equation can be written as \( \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \). Since the total degree of every term in both the numerator and denominator is equal to 2, this is a homogeneous differential equation. We solve it using the standard substitution method:
\[
y = vx \quad \Rightarrow \quad \frac{dy}{dx} = v + x\frac{dv}{dx}
\]
Step 1: Rewrite the differential equation in standard derivative form.
\[ \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \]
Step 2: Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).
Substituting these variables into our equation: \[ v + x\frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} \] Canceling \( x^2 \) from numerator and denominator gives: \[ v + x\frac{dv}{dx} = \frac{v}{1 + v^2} \]
Step 3: Separate the variables \( v \) and \( x \).
Isolate the term containing \( \frac{dv}{dx} \): \[ x\frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v(1 + v^2)}{1 + v^2} = \frac{v - v - v^3}{1 + v^2} = \frac{-v^3}{1 + v^2} \] Separating terms by moving all \( v \) components to the left and \( x \) components to the right: \[ \frac{1 + v^2}{v^3} \, dv = -\frac{1}{x} \, dx \] Splitting the fraction on the left: \[ \left( \frac{1}{v^3} + \frac{1}{v} \right) dv = -\frac{1}{x} \, dx \]
Step 4: Integrate both sides of the equation.
\[ \int \left( v^{-3} + \frac{1}{v} \right) dv = -\int \frac{1}{x} \, dx \] \[ \frac{v^{-2}}{-2} + \log|v| = -\log|x| + C \] \[ -\frac{1}{2v^2} + \log|v| + \log|x| = C \] Using logarithm properties \( \log|v| + \log|x| = \log|vx| \): \[ -\frac{1}{2v^2} + \log|vx| = C \]
Step 5: Substitute back \( v = \frac{y}{x} \) to obtain the final answer.
Since \( y = vx \), we can substitute \( vx = y \) and \( v = \frac{y}{x} \): \[ -\frac{1}{2\left(\frac{y}{x}\right)^2} + \log|y| = C \quad \Rightarrow \quad -\frac{x^2}{2y^2} + \log|y| = C \]
Step 1: Rewrite the differential equation in standard derivative form.
\[ \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \]
Step 2: Substitute \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).
Substituting these variables into our equation: \[ v + x\frac{dv}{dx} = \frac{x(vx)}{x^2 + (vx)^2} = \frac{vx^2}{x^2(1 + v^2)} \] Canceling \( x^2 \) from numerator and denominator gives: \[ v + x\frac{dv}{dx} = \frac{v}{1 + v^2} \]
Step 3: Separate the variables \( v \) and \( x \).
Isolate the term containing \( \frac{dv}{dx} \): \[ x\frac{dv}{dx} = \frac{v}{1 + v^2} - v = \frac{v - v(1 + v^2)}{1 + v^2} = \frac{v - v - v^3}{1 + v^2} = \frac{-v^3}{1 + v^2} \] Separating terms by moving all \( v \) components to the left and \( x \) components to the right: \[ \frac{1 + v^2}{v^3} \, dv = -\frac{1}{x} \, dx \] Splitting the fraction on the left: \[ \left( \frac{1}{v^3} + \frac{1}{v} \right) dv = -\frac{1}{x} \, dx \]
Step 4: Integrate both sides of the equation.
\[ \int \left( v^{-3} + \frac{1}{v} \right) dv = -\int \frac{1}{x} \, dx \] \[ \frac{v^{-2}}{-2} + \log|v| = -\log|x| + C \] \[ -\frac{1}{2v^2} + \log|v| + \log|x| = C \] Using logarithm properties \( \log|v| + \log|x| = \log|vx| \): \[ -\frac{1}{2v^2} + \log|vx| = C \]
Step 5: Substitute back \( v = \frac{y}{x} \) to obtain the final answer.
Since \( y = vx \), we can substitute \( vx = y \) and \( v = \frac{y}{x} \): \[ -\frac{1}{2\left(\frac{y}{x}\right)^2} + \log|y| = C \quad \Rightarrow \quad -\frac{x^2}{2y^2} + \log|y| = C \]
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