Question

Find the general solution of the differential equation \( 2x^2 \frac{dy}{dx} = y^2 + 2xy \).

Hint:

Always confirm an equation is homogeneous by checking if the total degree of every term is the same. The substitution \(y=vx\) always transforms a homogeneous equation into a variable-separable one.

Answer 1

Concept:
• A differential equation \( M(x,y)dx + N(x,y)dy = 0 \) is homogeneous if both functions are of the same degree.
• For homogeneous equations, use the substitution \( y = vx \), which implies \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).

Step 1: Rearrange and verify homogeneity
\[ \frac{dy}{dx} = \frac{y^2 + 2xy}{2x^2} = \frac{y^2}{2x^2} + \frac{2xy}{2x^2} = \frac{1}{2} \left( \frac{y}{x} \right)^2 + \frac{y}{x} \] Since the RHS is a function of \(y/x\), the equation is homogeneous.

Step 2: Substitute and separate variables
Let \(y = vx\). Then \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substitute into the equation: \[ v + x\frac{dv}{dx} = \frac{1}{2}v^2 + v \] Subtract \(v\) from both sides: \[ x\frac{dv}{dx} = \frac{v^2}{2} \] Separate variables: \[ \frac{2}{v^2} \, dv = \frac{dx}{x} \]

Step 3: Integrate both sides
\[ 2 \int v^{-2} \, dv = \int \frac{1}{x} \, dx \] \[ 2 \left( \frac{v^{-1}}{-1} \right) = \log|x| + C \] \[ -\frac{2}{v} = \log|x| + C \]

Step 4: Substitute back for \(y\)
Since \(v = y/x\), then \(1/v = x/y\). \[ -\frac{2x}{y} = \log|x| + C \] Or, expressed for \(y\): \[ y = \frac{-2x}{\log|x| + C} \]