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 rewriting it as dy/dx = -y²/(x² - xy + y²), then use the substitution y = vx to reduce it to a separable form.
<p><strong>Step 1:</strong> Rewrite the equation as: y² dx + (x² - xy + y²) dy = 0</p><p>Rearranging: dy/dx = -y²/(x² - xy + y²)</p><p><strong>Step 2:</strong> Check if homogeneous: Both numerator and denominator are homogeneous of degree 2, so use substitution y = vx, where dy = v dx + x dv</p><p><strong>Step 3:</strong> Substitute into dy/dx = -y²/(x² - xy + y²):</p><p>v + x(dv/dx) = -(vx)²/(x² - x(vx) + (vx)²) = -v²x²/(x²(1 - v + v²))</p><p>v + x(dv/dx) = -v²/(1 - v + v²)</p><p><strong>Step 4:</strong> Separate variables:</p><p>x(dv/dx) = -v²/(1 - v + v²) - v = -(v² + v(1 - v + v²))/(1 - v + v²) = -(v + v³)/(1 - v + v²)</p><p>x(dv/dx) = -v(1 + v²)/(1 - v + v²)</p><p><strong>Step 5:</strong> Separate and integrate:</p><p>∫(1 - v + v²)/(v(1 + v²)) dv = -∫dx/x</p><p>After partial fraction decomposition and integration: ln|v| - (1/2)ln(1 + v²) + arctan(v) = -ln|x| + C</p><p><strong>Step 6:</strong> Substitute v = y/x back and simplify to get the implicit solution in the form of option A</p><p>∴ Answer: A</p>
Correct Answer: A