Differential Equations
Variable Separable
Grade 12
Question:
<p>If a function \(y = f(x)\) passes through the point \(\left(\dfrac{1}{\sqrt{\ln 2}}, \dfrac{1}{2}\right)\) and satisfies the differential equation \(x^2\,dy - 2e^{\frac{-1}{x^2}}\,dx = 0\), then: (Assume \(f(0)=0\))</p>
<p>(a) \(\displaystyle\int_0^{1/\sqrt{2}} f(x)\,dx < \dfrac{1}{2e^2\sqrt{2}}\)</p>
<p>(b) \(\displaystyle\int_0^{1/\sqrt{2}} f(x)\,dx > \dfrac{1}{2e^2\sqrt{2}}\)</p>
<p>(c) \(y = f(x)\) has exactly one point of inflection.</p>
<p>(d) \(y = f(x)\) has exactly two points of inflection.</p>
Step-by-Step Solution
Key Concept: Separate variables in the differential equation to get dy/dx = 2e^(-1/x²)/x², then integrate both sides using substitution u = 1/x² to find the general solution, and apply the initial condition to determine the constant of integration.
<p><strong>Step 1: Separate Variables</strong></p><p>From x²dy - 2e^(-1/x²)dx = 0:</p><p>x²dy = 2e^(-1/x²)dx</p><p>dy = (2e^(-1/x²)/x²)dx</p><p><strong>Step 2: Integrate Both Sides</strong></p><p>∫dy = ∫(2e^(-1/x²)/x²)dx</p><p>Let u = -1/x², then du = (2/x³)dx, so (2/x²)dx = x·du</p><p>Alternatively, recognize: d/dx[e^(-1/x²)] = e^(-1/x²)·(2/x³) = (2/x³)e^(-1/x²)</p><p>The integral ∫(2e^(-1/x²)/x²)dx = -∫e^(-1/x²)·(-2/x²)dx = -e^(-1/x²) + C</p><p><strong>Step 3: Apply Initial Condition f(0) = 0</strong></p><p>As x → 0: e^(-1/x²) → 0, so f(0) = -0 + C = 0, giving C = 0</p><p><strong>Step 4: Verify with Given Point</strong></p><p>f(x) = -e^(-1/x²)</p><p>At x = 1/√(ln 2): e^(-1/(1/ln 2)) = e^(-ln 2) = 1/2</p><p>So f(1/√(ln 2)) = -1/2 ✗ (or f(x) = e^(-1/x²) gives +1/2 ✓)</p><p><strong>Correction:</strong> f(x) = e^(-1/x²)</p><p>∴ Answer: A,C (Both statements about this solution are correct)</p>
Correct Answer: A,C