Differential Equations
Formation and solution of differential equations
Grade 12
Question:
<p>The equation of a curve passing through (1, 0) for which the product of the abscissa of a point \(P\) and the intercept made by a normal at \(P\) on the \(x\)-axis equals twice the square of the radius vector of the point \(P\), is</p>
<p>(a) \(x^2 + y^2 = Cx^4\)</p>
<p>(b) \(x^2 + y^2 = 2x^4\)</p>
<p>(c) \(x^2 - y^2 = 4x^4\)</p>
<p>(d) \(x^2 - y^2 = x^4\)</p>
Step-by-Step Solution
Key Concept: Set up the differential equation by interpreting the geometric condition: the x-intercept of the normal at P(x,y) is x + y(dy/dx), and the radius vector squared is x² + y². The product condition xy(x + y·dy/dx) = 2(x² + y²) transforms into a solvable differential equation.
<p><strong>Step 1:</strong> For point P(x,y), the normal has slope −1/(dy/dx). Normal equation: Y − y = −(dx/dy)(X − x). Setting Y = 0 gives x-intercept at X = x + y(dy/dx).</p><p><strong>Step 2:</strong> Given condition: x·[x + y(dy/dx)] = 2(x² + y²). This simplifies to: xy(dy/dx) = 2y² + x².</p><p><strong>Step 3:</strong> Rearrange: dy/dx = (2y² + x²)/(xy). This is homogeneous; substitute y = vx, so dy/dx = v + x(dv/dx).</p><p><strong>Step 4:</strong> v + x(dv/dx) = (2v² + 1)/v. Therefore: x(dv/dx) = (v² + 1)/v. Separating variables: v·dv/(v² + 1) = dx/x.</p><p><strong>Step 5:</strong> Integrate: ½ln(v² + 1) = ln|x| + C. Using boundary condition (1,0): v = 0 at x = 1 gives C = −½ln(1) = 0.</p><p><strong>Step 6:</strong> Thus ln(v² + 1) = 2ln|x|, so v² + 1 = x². With v = y/x: (y/x)² + 1 = x², which gives x² − y² = 1.</p><p>∴ Answer: x² − y² = 1 (or equivalent form)</p>
Correct Answer: A