Differential Equations
Variable Separable / Homogeneous Equations
Grade 12
Question:
<p>Given \((y^2 - x^3)\,dx - xy\,dy = 0\). Find the solution of the differential equation.</p>
<p>\(-\dfrac{1}{2}\dfrac{y^2}{x^2} = x + c\)</p>
<p>\(\dfrac{1}{2}\dfrac{y^2}{x^2} = x + c\)</p>
<p>\(-\dfrac{y^2}{x^2} = x + c\)</p>
<p>\(\dfrac{y^2}{x^2} = x + c\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a homogeneous differential equation by rewriting it in the form M(x,y)dx + N(x,y)dy = 0, then use the substitution y = vx to reduce it to a separable equation.
<p><strong>Step 1:</strong> Write the equation in standard form: (y² - x³)dx - xy dy = 0</p><p><strong>Step 2:</strong> Check homogeneity. Both M(x,y) = y² - x³ and N(x,y) = -xy are homogeneous of degree 2, confirming this is a homogeneous DE.</p><p><strong>Step 3:</strong> Use substitution y = vx, so dy = v dx + x dv</p><p><strong>Step 4:</strong> Substitute into the equation:<br/>(v²x² - x³)dx - x(vx)(v dx + x dv) = 0<br/>x²(v² - x)dx - vx²(v dx + x dv) = 0</p><p><strong>Step 5:</strong> Simplify:<br/>x²(v² - x)dx - v²x² dx - vx³ dv = 0<br/>-x³ dx - vx³ dv = 0<br/>-dx/x - v dv = 0</p><p><strong>Step 6:</strong> Integrate both sides:<br/>-ln|x| - v²/2 = C<br/>2ln|x| + v² = -2C</p><p><strong>Step 7:</strong> Substitute back v = y/x:<br/>2ln|x| + y²/x² = C₁<br/>∴ <strong>x²(y² + 2x² ln|x|) = Cx²</strong> or <strong>y² + 2x² ln|x| = C</strong></p>
Correct Answer: A