Bernoulli Differential Equation

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 $$x \to \frac{\pi}{2}$$:

$$\lim_{x \to \pi/2} \left(\frac{x^3}{y(x)^3}\right) = -3 \sin\left(\frac{\pi}{2}\right) + 5$$

$$= -3(1) + 5 = 2$$