Question

Let \(y = y(x)\) be the solution of the differential equation \(\sec(x) \frac{dy}{dx} + [2(1 - x) \tan(x) + x(2 - x)] = 0\) such that \(y(0) = 2\). Then \(y(2)\) is equal to:

• A. 2
• B. 2\{1 - \sin(2)\}
• C. 2\{\sin(2) + 1\}
• D. 1

Answer 1

The Correct Option is A

Solution: We start with the differential equation:

\(\sec x \, dy + \{ 2(1 - x) \tan x + x(2 - x) \} \, dx = 0\)

Divide by \(\sec x\) to simplify:

\(\dfrac{dy}{dx} = -\{ 2(1 - x) \sin x + x(2 - x) \cos x \}\)

Integrate both sides:

\(y(x) = - \int \{ 2(1 - x) \sin x + x(2 - x) \cos x \} \, dx + C\)

Separate the integrals:

\(y(x) = - \int 2(1 - x) \sin x \, dx - \int x(2 - x) \cos x \, dx + C\)

Calculate each integral:

\(y(x) = (x^2 - 2x) \sin x + C\)

Using the initial condition \(y(0) = 2\):

\(y(0) = 0 + C \Rightarrow C = 2\)

Thus,

\(y(x) = (x^2 - 2x) \sin x + 2\)

Finally, substituting \(x = 2\):

\(y(2) = (2^2 - 2 \times 2) \sin 2 + 2 = 2\)