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

Answer: (A)

2

Answer 2

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 $x = 2$:

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