Question:
Find the general solution of the differential equation \( (y^2 - x^2)dx = 2xy dy \)
Find the general solution of the differential equation \( (y^2 - x^2)dx = 2xy dy \)
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A differential equation \( Mdx + Ndy = 0 \) is homogeneous if \( M \) and \( N \) are of the same degree.
For homogeneous equations, the substitution \( y = vx \) always reduces it to variable separable form.
For homogeneous equations, the substitution \( y = vx \) always reduces it to variable separable form.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Homogeneous Differential Equations: Equations where the total degree of every term is the same.
• Method of substitution: Put \( y = vx \).
Step 1: Identify as homogeneous and substitute
Rearranging the equation:
\[ \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \] Put \( y = vx \), then \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). \[ v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2x(vx)} = \frac{v^2 - 1}{2v} \]
Step 2: Separate the variables
\[ x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = -\frac{1 + v^2}{2v} \] \[ \frac{2v}{1 + v^2} dv = -\frac{1}{x} dx \]
Step 3: Integrate and substitute back
\[ \int \frac{2v}{1 + v^2} dv = -\int \frac{dx}{x} \] \[ \log(1 + v^2) = -\log x + \log C = \log(\frac{C}{x}) \] \[ 1 + v^2 = \frac{C}{x} \implies 1 + \frac{y^2}{x^2} = \frac{C}{x} \] \[ \frac{x^2 + y^2}{x^2} = \frac{C}{x} \implies x^2 + y^2 = Cx \]
• Homogeneous Differential Equations: Equations where the total degree of every term is the same.
• Method of substitution: Put \( y = vx \).
Step 1: Identify as homogeneous and substitute
Rearranging the equation:
\[ \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \] Put \( y = vx \), then \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). \[ v + x \frac{dv}{dx} = \frac{v^2 x^2 - x^2}{2x(vx)} = \frac{v^2 - 1}{2v} \]
Step 2: Separate the variables
\[ x \frac{dv}{dx} = \frac{v^2 - 1}{2v} - v = \frac{v^2 - 1 - 2v^2}{2v} = -\frac{1 + v^2}{2v} \] \[ \frac{2v}{1 + v^2} dv = -\frac{1}{x} dx \]
Step 3: Integrate and substitute back
\[ \int \frac{2v}{1 + v^2} dv = -\int \frac{dx}{x} \] \[ \log(1 + v^2) = -\log x + \log C = \log(\frac{C}{x}) \] \[ 1 + v^2 = \frac{C}{x} \implies 1 + \frac{y^2}{x^2} = \frac{C}{x} \] \[ \frac{x^2 + y^2}{x^2} = \frac{C}{x} \implies x^2 + y^2 = Cx \]
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