Differential Equations
Homogeneous Differential Equations
Grade 12
Question:
<p>The solution of the differential equation \(x\dfrac{dy}{dx} = y(\log y - \log x + 1)\) is:</p>
<p>(1) \(y = xe^{cx}\)</p>
<p>(2) \(\log\left(\dfrac{y}{x}\right) = cx\)</p>
<p>(3) \(y = e^{cx} \cdot x\)</p>
<p>(4) Both (2) and (3)</p>
Step-by-Step Solution
Key Concept: Recognize this as a homogeneous differential equation by rewriting it as x(dy/dx) = y(log(y/x) + 1), then use the substitution v = y/x to convert it into a separable form.
<p><strong>Step 1:</strong> Rewrite the given equation</p><p>x(dy/dx) = y(log y - log x + 1) = y(log(y/x) + 1)</p><p><strong>Step 2:</strong> Recognize as homogeneous equation. Use substitution v = y/x, so y = vx and dy/dx = v + x(dv/dx)</p><p><strong>Step 3:</strong> Substitute into the equation:</p><p>x(v + x(dv/dx)) = vx(log v + 1)</p><p>xv + x²(dv/dx) = vx log v + vx</p><p>x²(dv/dx) = vx log v</p><p><strong>Step 4:</strong> Separate variables:</p><p>(dv)/(v log v) = dx/x</p><p><strong>Step 5:</strong> Integrate both sides:</p><p>∫(dv)/(v log v) = ∫dx/x</p><p>log(log v) = log x + C₁</p><p>log(log v) = log(Cx) where C = e^(C₁)</p><p><strong>Step 6:</strong> Solve for y:</p><p>log v = Cx</p><p>log(y/x) = Cx</p><p><p>∴ Answer: <strong>log(y/x) = Cx</strong> or equivalently <strong>y = xe^(Cx)</strong></p>
Correct Answer: C