Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
A function $y = f(x)$ satisfying the differential equation $\frac{dy}{dx} - \sin x - y\cos x + \sin^2 x \cdot \frac{\sin^2 x}{x^2} = 0$ is such that, $y\left(\frac{\pi}{2}\right) = \frac{2}{\pi}$, then the statement which is correct?
$\lim_{x \to 0^+} f(x) = 1$
$\int_0^{\pi/2} f(x)dx$ is less than $\frac{\pi}{2}$
$\int_0^{\pi/2} f(x)dx$ is greater than unity
$f(x)$ is an odd function
Step-by-Step Solution
Key Concept: Identify the integrating factor for a linear first-order ODE and apply boundary conditions to find the particular solution.
The differential equation $\frac{dy}{dx} - y\cot x = \frac{\sin x}{x^2}$ is linear with integrating factor $e^{-\cot x\,dx} = e^{-\log\sin x} = \frac{1}{\sin x}$. Multiplying through and integrating gives $y \cdot \frac{1}{\sin x} = -\frac{1}{x} + C$. Using the boundary condition $y(\frac{\pi}{2}) = \frac{2}{\pi}$ yields $C = 0$, so $y = \frac{\sin x}{x}$. The integral $I = \int_0^{\pi/2} \frac{\sin x}{x}\,dx$ is positive since $\frac{\sin x}{x}$ is decreasing for $x > 0$, making $f(x) > f(\frac{\pi}{2})$ over most of $[0, \frac{\pi}{2})$.
Correct Answer: 1,2,3