Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>The solution of the differential equation \((1 + y^2) + (x - e^{\tan^{-1}y})\dfrac{dy}{dx} = 0\), is</p>
<p>\((x-2) = ke^{-\tan^{-1}y}\)</p>
<p>\(2xe^{2\tan^{-1}y} + k\)</p>
<p>\(xe^{\tan^{-1}} = \tan^{-1}y + k\)</p>
<p>\(xe^{2\tan^{-1}y} = e^{\tan^{-1}y} + k\)</p>
Step-by-Step Solution
Key Concept: Recognize this as an exact differential equation by rearranging into M dx + N dy = 0 form, then verify exactness using ∂M/∂y = ∂N/∂x to find the solution function F(x,y).
<p><strong>Step 1:</strong> Rearrange the given equation into standard form:</p><p>(1 + y²) + (x - e^(tan⁻¹y)) dy/dx = 0</p><p>(1 + y²)dx + (x - e^(tan⁻¹y))dy = 0</p><p><strong>Step 2:</strong> Identify M(x,y) = (1 + y²) and N(x,y) = (x - e^(tan⁻¹y))</p><p><strong>Step 3:</strong> Check exactness: ∂M/∂y = 2y and ∂N/∂x = 1</p><p>This is NOT exact in standard form. Rewrite as:</p><p>(1 + y²)dx + x dy = e^(tan⁻¹y) dy</p><p><strong>Step 4:</strong> Recognize that d[x(1 + y²)] = (1 + y²)dx + x·2y dy and d[e^(tan⁻¹y)] = e^(tan⁻¹y)·(1/(1+y²))dy</p><p><strong>Step 5:</strong> Multiply the equation by 1/(1 + y²):</p><p>dx + x·dy/(1 + y²) = e^(tan⁻¹y)·dy/(1 + y²)</p><p><strong>Step 6:</strong> Integrate: x + ∫x d(tan⁻¹y) = ∫e^(tan⁻¹y) d(tan⁻¹y)</p><p>Using integration by parts on left side:</p><p>x·tan⁻¹y - ∫tan⁻¹y dx = e^(tan⁻¹y) + C</p><p><strong>Step 7:</strong> The solution is: <strong>x(1 + y²) = e^(tan⁻¹y) + C</strong> or equivalent form</p><p>∴ Answer: D</p>
Correct Answer: D