Differential Equations
Linear Differential Equations
Grade 12
Question:
<p>Solution of the differential equation \(\cos x\,dy = y(\sin x - y)\,dx,\; 0 < x < \dfrac{\pi}{2}\) is</p>
<p>\(y\sec x = \tan x + c\)</p>
<p>\(y\tan x = \sec x + c\)</p>
<p>\(\tan x = (\sec x + c)y\)</p>
<p>\(\sec x = (\tan x + c)y\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a Bernoulli equation by dividing by y² and substituting v = 1/y to convert it into a linear first-order DE. The presence of y² term coupled with y and constant terms signals Bernoulli's form.
<p><strong>Step 1:</strong> Rewrite the equation: cos x dy = y(sin x - y) dx</p><p>⟹ cos x dy = y sin x dx - y² dx</p><p>Dividing by y² cos x: (1/y²)(dy/dx) = (sin x/cos x)(1/y) - 1</p><p><strong>Step 2:</strong> This is Bernoulli's form. Let v = 1/y, so dv/dx = -(1/y²)(dy/dx)</p><p>Substituting: -dv/dx = (tan x)v - 1</p><p>⟹ dv/dx + (tan x)v = 1</p><p><strong>Step 3:</strong> This is linear in v. Integrating factor: IF = e^(∫tan x dx) = e^(ln|sec x|) = sec x</p><p><strong>Step 4:</strong> Multiply by IF: sec x(dv/dx) + sec x tan x v = sec x</p><p>⟹ d/dx(v sec x) = sec x</p><p><strong>Step 5:</strong> Integrate: v sec x = ∫sec x dx = ln|sec x + tan x| + C</p><p><strong>Step 6:</strong> Since v = 1/y:</p><p>∴ sec x/y = ln|sec x + tan x| + C</p><p>or <strong>y cos x = 1/[ln|sec x + tan x| + C]</strong></p>
Correct Answer: D