Differential Equations
Homogeneous Differential Equations
Grade 12
Question:
<p><strong>98.</strong> The solution of the differential equation \(y^2\, dx + (x^2 - xy + y^2)\, dy = 0\) is:</p><p>[<strong>Note:</strong> Where \(C\) is constant of integration.]</p>
<p>\(\tan^{-1}\!\left(\dfrac{x}{y}\right) + \ln y + C = 0\)</p>
<p>\(2\tan^{-1}\!\left(\dfrac{x}{y}\right) + \ln y + C = 0\)</p>
<p>\(\ln\!\left(y + \sqrt{x^2 + y^2}\right) + \ln y + C = 0\)</p>
<p>\(\ln\!\left(y + \sqrt{x^2 + y^2}\right) + C = 0\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a homogeneous differential equation by checking if M(λx,λy) = λⁿM(x,y). Use the substitution y = vx to convert it into a separable form in variables v and x.
<p><strong>Step 1:</strong> Verify homogeneity. The equation y²dx + (x² - xy + y²)dy = 0 has M(x,y) = y² and N(x,y) = x² - xy + y². Both are degree 2 polynomials, confirming homogeneity.</p><p><strong>Step 2:</strong> Substitute y = vx, so dy = vdx + xdv. The equation becomes: v²x²dx + (x² - vx² + v²x²)(vdx + xdv) = 0</p><p><strong>Step 3:</strong> Expand: v²x²dx + x²(1 - v + v²)(vdx + xdv) = 0. Simplifying: v²x²dx + vx²(1 - v + v²)dx + x³(1 - v + v²)dv = 0</p><p><strong>Step 4:</strong> Collect dx terms: [v² + v(1 - v + v²)]x²dx + x³(1 - v + v²)dv = 0, which gives [v + v²]x²dx + x³(1 - v + v²)dv = 0</p><p><strong>Step 5:</strong> Separate variables: [v(1 + v)]x²dx + x³(1 - v + v²)dv = 0 → dx/x + (1 - v + v²)/[v(1 + v)]dv = 0</p><p><strong>Step 6:</strong> Using partial fractions and integrating: ln|x| + ln|v| - ln|1 + v| + C₁ = 0, which simplifies to xv/(1 + v) = C or xy/(x + y) = C</p><p>∴ Answer: A</p>
Correct Answer: A