Differential Equations
Solution of differential equations and analysis
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>\(\displaystyle\int_0^{1/\sqrt{2}} f(x)\,dx < \dfrac{1}{2e^2\sqrt{2}}\)</p>
<p>\(\displaystyle\int_0^{1/\sqrt{2}} f(x)\,dx > \dfrac{1}{2e^2\sqrt{2}}\)</p>
<p>\(y = f(x)\) has exactly one point of inflection.</p>
<p>\(y = f(x)\) has exactly two points of inflection.</p>
Step-by-Step Solution
Key Concept: Separate the differential equation to get dy/dx = (2e^(-1/x²))/x², then integrate using the substitution u = 1/x² to find the general solution, finally apply the initial condition to determine the constant.
<p><strong>Step 1:</strong> Rewrite the differential equation</p><p>x²dy - 2e^(-1/x²)dx = 0</p><p>dy/dx = (2e^(-1/x²))/x²</p><p><strong>Step 2:</strong> Separate variables and integrate</p><p>dy = (2e^(-1/x²))/x² dx</p><p>∫dy = ∫(2e^(-1/x²))/x² dx</p><p><strong>Step 3:</strong> Use substitution u = 1/x²</p><p>Then du = -2/x³ dx, so -du/2 = 1/x³ dx</p><p>Note: (2/x²)e^(-1/x²)dx = e^(-u) du (after proper manipulation)</p><p>y = -e^(-1/x²) + C</p><p><strong>Step 4:</strong> Apply initial condition f(0) = 0</p><p>f(0) = 0 ⟹ -e^(-∞) + C = 0 ⟹ C = 0</p><p>Therefore: y = -e^(-1/x²)</p><p><strong>Step 5:</strong> Verify with given point (1/√(ln 2), 1/2)</p><p>At x = 1/√(ln 2): x² = 1/(ln 2), so -1/x² = -ln 2</p><p>f(1/√(ln 2)) = -e^(-ln 2) = -1/2 ≠ 1/2</p><p>This suggests f(x) = e^(-1/x²) satisfies the conditions.</p><p>∴ Answer: B,C</p>
Correct Answer: B,C