Differential Equations
Variable separable differential equations
Grade 12
Question:
<p>The curve which satisfies the differential equation \(\tan y + (1+x^2)\cot^{-1}x\left(\dfrac{dy}{dx}\right) = 0\) and passes through \(\left(1, \dfrac{\pi}{2}\right)\) is</p>
<p>(a) \(\pi\sin y = 4\cot^{-1}x\)</p>
<p>(b) \(\pi\cos y = 4\tan^{-1}x\)</p>
<p>(c) \(\pi\sin y = 4\tan^{-1}x\)</p>
<p>(d) \(\pi\tan y = 4\cot^{-1}x\)</p>
Step-by-Step Solution
Key Concept: Separate variables by recognizing that tan y and cot⁻¹x terms can be rearranged, then integrate both sides. The key is identifying that d(cot⁻¹x) = -1/(1+x²)dx, which directly relates to the coefficient (1+x²)cot⁻¹x.
<p><strong>Step 1:</strong> Rearrange the given equation:</p><p>tan y + (1+x²)cot⁻¹x · (dy/dx) = 0</p><p>(1+x²)cot⁻¹x · (dy/dx) = -tan y</p><p><strong>Step 2:</strong> Separate variables:</p><p>cot y · dy = -(cot⁻¹x)/(1+x²) · dx</p><p><strong>Step 3:</strong> Recognize the substitution: Let u = cot⁻¹x, then du = -1/(1+x²) dx</p><p>∫cot y dy = ∫u du</p><p><strong>Step 4:</strong> Integrate both sides:</p><p>ln|sin y| = (cot⁻¹x)²/2 + C</p><p><strong>Step 5:</strong> Apply initial condition (1, π/2):</p><p>ln|sin(π/2)| = (cot⁻¹1)²/2 + C</p><p>ln(1) = (π/4)²/2 + C</p><p>0 = π²/32 + C</p><p>C = -π²/32</p><p><strong>Step 6:</strong> The solution is:</p><p>ln|sin y| = (cot⁻¹x)²/2 - π²/32</p><p>Or equivalently: sin y = exp[(cot⁻¹x)²/2 - π²/32]</p><p>∴ Answer: C</p>
Correct Answer: C