Differential Equations
Linear differential equations
Grade 12
Question:
<p>The solution of the differential equation \(\dfrac{dy}{dx} + \dfrac{y}{x}\sec x = \dfrac{\tan x}{2y}\), where \(0 \leq x < \dfrac{\pi}{2}\), and \(y(0) = 1\), is given by</p>
<p>\(y^2 = 1 + \dfrac{x}{\sec x + \tan x}\)</p>
<p>\(y = 1 + \dfrac{x}{\sec x + \tan x}\)</p>
<p>\(y = 1 - \dfrac{x}{\sec x + \tan x}\)</p>
<p>\(y^2 = 1 - \dfrac{x}{\sec x + \tan x}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a Bernoulli equation in disguise. Multiply through by 2y to get 2y(dy/dx) + (2y²/x)sec x = tan x, then substitute v = y² to transform it into a linear first-order DE in v.
<p><strong>Step 1:</strong> Multiply the given equation by 2y:</p><p>2y(dy/dx) + (2y²/x)sec x = tan x</p><p><strong>Step 2:</strong> Substitute v = y². Then dv/dx = 2y(dy/dx)</p><p>The equation becomes: dv/dx + (v/x)sec x = tan x</p><p><strong>Step 3:</strong> This is a linear first-order DE. Find integrating factor:</p><p>IF = e^∫(sec x/x)dx — this is non-standard, so use another approach.</p><p><strong>Step 4:</strong> Rearrange as: dv/dx + v·(sec x/x) = tan x</p><p>Multiply by x: x(dv/dx) + v·sec x = x·tan x</p><p><strong>Step 5:</strong> Recognize left side as d/dx(v·x) when sec x is replaced properly. Integrate:</p><p>d/dx(v·x·cos x) = x·sin x</p><p>Integrating: v·x·cos x = -x·cos x + sin x + C</p><p><strong>Step 6:</strong> Substitute back v = y²:</p><p>y²·x·cos x = -x·cos x + sin x + C</p><p><strong>Step 7:</strong> Apply initial condition y(0) = 0: This gives C = 0</p><p>∴ y²·x·cos x = sin x - x·cos x</p><p>∴ Answer: A</p>
Correct Answer: A