Step 1: Convert the Bernoulli equation to linear form
The given differential equation is:
$$x^3 \frac{dy}{dx} - x^2 y = y^4 \cos x$$
Divide the entire equation by $$y^4$$ to separate the variables containing $$y$$:
$$x^3 y^{-4} \frac{dy}{dx} - x^2 y^{-3} = \cos x$$
Step 2: Use substitution
Let $$u = y^{-3}$$. Differentiating both sides with respect to $$x$$ gives:
$$\frac{du}{dx} = -3y^{-4} \frac{dy}{dx} \implies y^{-4} \frac{dy}{dx} = -\frac{1}{3} \frac{du}{dx}$$
Substitute these into the equation:
$$x^3 \left(-\frac{1}{3} \frac{du}{dx}\right) - x^2 u = \cos x$$
Multiply the entire equation by $$-\frac{3}{x^3}$$ to write it in standard linear form:
$$\frac{du}{dx} + \frac{3}{x} u = -\frac{3 \cos x}{x^3}$$
Step 3: Find the integrating factor (I.F.)
$$\text{I.F.} = e^{\int \frac{3}{x} \, dx} = e^{3 \ln x} = x^3$$
Multiplying the differential equation by the $$\text{I.F.}$$:
$$\frac{d}{dx}\left(u \cdot x^3\right) = -\frac{3 \cos x}{x^3} \cdot x^3$$
$$\frac{d}{dx}\left(u \cdot x^3\right) = -3 \cos x$$
Step 4: Integrate to find the general solution
Integrate both sides with respect to $$x$$:
$$u \cdot x^3 = \int -3 \cos x \, dx + C$$
$$u \cdot x^3 = -3 \sin x + C$$
Substitute back $$u = \frac{1}{y^3}$$:
$$\frac{x^3}{y^3} = -3 \sin x + C$$
Step 5: Apply the given boundary condition
We are given $$\lim_{x \to \pi} y(x) = \left(\frac{5}{\pi^3}\right)^{-1/3}$$, which means:
$$y(\pi)^3 = \frac{\pi^3}{5}$$
Substitute $$x = \pi$$ and $$y^3 = \frac{\pi^3}{5}$$ into the general solution:
$$\frac{\pi^3}{\frac{\pi^3}{5}} = -3 \sin(\pi) + C$$
$$5 = -3(0) + C \implies C = 5$$
So, the exact relation is:
$$\frac{x^3}{y^3} = -3 \sin x + 5$$
Step 6: Evaluate the required limit
Now, find the value as $$x \to \frac{\pi}{2}$$:
$$\lim_{x \to \pi/2} \left(\frac{x^3}{y(x)^3}\right) = -3 \sin\left(\frac{\pi}{2}\right) + 5$$
$$= -3(1) + 5 = 2$$
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