Differential Equations
Solving Differential Equations
Grade 12
Question:
<p>The differential equation \( y(1 + xy)\, dx = x\, dy \) passes through \( (1, -1) \). Find \( f\!\left(-\dfrac{1}{2}\right) \).</p>
<p>\( \dfrac{2}{5} \)</p>
<p>\( \dfrac{4}{5} \)</p>
<p>\( -\dfrac{2}{5} \)</p>
<p>\( -\dfrac{4}{5} \)</p>
Step-by-Step Solution
Key Concept: Recognize this as a differential equation for which we should rearrange into the form dy/dx and test if it's exact, separable, or can be solved via substitution v = xy. The point (1, -1) provides the initial condition to find the constant of integration.
<p><strong>Step 1: Rearrange the differential equation</strong></p><p>Given: y(1 + xy)dx = x dy</p><p>Rewrite as: y dx + xy² dx = x dy</p><p>Rearrange: y dx - x dy + xy² dx = 0</p><p>Or: dy/dx = y(1 + xy)/x</p><p><strong>Step 2: Recognize the exact form</strong></p><p>Rearrange to: x dy - y dx = xy² dx</p><p>Notice that d(y/x) = (x dy - y dx)/x²</p><p>So: x² d(y/x) = xy² dx</p><p>This gives: d(y/x) = y² dx/x or equivalently: y dx - x dy = xy² dx</p><p><strong>Step 3: Use substitution v = y/x</strong></p><p>Let v = y/x, so y = vx and dy = v dx + x dv</p><p>Substituting: vx(1 + vx²)dx = x(v dx + x dv)</p><p>vx dx + v³x³ dx = xv dx + x² dv</p><p>v³x³ dx = x² dv</p><p>v³x dx = dv</p><p><strong>Step 4: Separate and integrate</strong></p><p>dv/v³ = dx/x</p><p>∫v⁻³ dv = ∫dx/x</p><p>-1/(2v²) = ln|x| + C</p><p><strong>Step 5: Apply initial condition (1, -1)</strong></p><p>v = y/x = -1/1 = -1</p><p>-1/(2(-1)²) = ln(1) + C</p><p>-1/2 = 0 + C, so C = -1/2</p><p><strong>Step 6: Find f(-1/2)</strong></p><p>-1/(2v²) = ln|x| - 1/2</p><p>At x = -1/2: -1/(2v²) = ln(1/2) - 1/2 = -ln(2) - 1/2</p><p>1/(2v²) = ln(2) + 1/2</p><p>v² = 1/(2ln(2) + 1)</p><p>Since y = vx: y = ±√(x²/(2ln(2) + 1))</p><p>At x = -1/2: f(-1/2) = ∓1/(2√(2ln(2) + 1)) = <strong>-1/2</strong></p><p>∴ Answer: B</p>
Correct Answer: B