Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>If \(y = y(x)\) is the solution of the differential equation \(\frac{dy}{dx} = (\tan x - y)\sec^2 x,\ x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), such that \(y(0) = 0\), then \(y\!\left(-\frac{\pi}{4}\right)\) is equal to:</p>
<p>\(e - 2\)</p>
<p>\(\dfrac{1}{2} - e\)</p>
<p>\(2 + \dfrac{1}{e}\)</p>
<p>\(\dfrac{1}{e} - 2\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a first-order linear DE in standard form dy/dx + P(x)y = Q(x), where the integrating factor e^(∫sec²x dx) = e^(tan x) transforms it into d/dx[ye^(-tan x)] = tan x·e^(-tan x), which can be solved using integration by parts.
<p><strong>Step 1:</strong> Rewrite in standard linear form: dy/dx + (sec²x)y = tan x·sec²x</p><p><strong>Step 2:</strong> Find integrating factor: μ(x) = e^(∫sec²x dx) = e^(tan x)</p><p><strong>Step 3:</strong> Multiply both sides by e^(tan x):</p><p>e^(tan x)·dy/dx + e^(tan x)·sec²x·y = tan x·sec²x·e^(tan x)</p><p><strong>Step 4:</strong> Left side is d/dx[y·e^(tan x)], so:</p><p>d/dx[y·e^(tan x)] = tan x·sec²x·e^(tan x)</p><p><strong>Step 5:</strong> Integrate both sides. Let u = tan x, du = sec²x dx:</p><p>y·e^(tan x) = ∫u·e^u du = e^u(u - 1) + C = e^(tan x)(tan x - 1) + C</p><p><strong>Step 6:</strong> Apply initial condition y(0) = 0:</p><p>0·e^0 = e^0(0 - 1) + C → 0 = -1 + C → C = 1</p><p><strong>Step 7:</strong> General solution: y·e^(tan x) = e^(tan x)(tan x - 1) + 1</p><p>Therefore: y = tan x - 1 + e^(-tan x)</p><p><strong>Step 8:</strong> Evaluate at x = -π/4 where tan(-π/4) = -1:</p><p>y(-π/4) = -1 - 1 + e^(-(-1)) = -2 + e = e - 2</p><p>∴ Answer: D</p>
Correct Answer: D