Differential Equations Question - None

Question

If $y = f(x)$ satisfies the differential equation $\sin x \dfrac{dy}{dx} + 2y\cos x = 8$ with $f(\pi/2) = 8$, then the minimum value of $f(x)$ is:

Accepted Answer

Step 1: Rewrite the equation by dividing throughout by sin x:

dy/dx + (2cos x/sin x)y = 8/sin x

dy/dx + 2cot(x)·y = 8csc(x)

Step 2: Identify P(x) = 2cot(x). Find integrating factor:

μ(x) = e^∫2cot(x)dx = e^(2ln|sin x|) = sin²x

Step 3: Multiply the equation by μ(x) = sin²x:

sin²x·dy/dx + 2sin x cos x·y = 8sin x

Step 4: The left side is d/dx(sin²x·y):

d/dx(sin²x·y) = 8sin x

Step 5: Integrate both sides:

sin²x·y = ∫8sin x dx = -8cos x + C

y = (-8cos x + C)/sin²x

Step 6: Apply initial condition f(π/2) = 8:

8 = (-8·0 + C)/1 ⟹ C = 8

y = (8 - 8cos x)/sin²x = 8(1 - cos x)/sin²x

Step 7: Find minimum. Use 1 - cos x = 2sin²(x/2) and sin²x = 4sin²(x/2)cos²(x/2):

y = 8·2sin²(x/2)/(4sin²(x/2)cos²(x/2)) = 4/cos²(x/2)

Minimum occurs when cos²(x/2) is maximum = 1 at x = 0 (within domain)

∴ Minimum value = 4

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