Rewrite the Bernoulli differential equation in standard form.

The given equation is $\frac{dy}{dx} - \frac{y}{x} = y^2 \cos(x)$.
This is a Bernoulli equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$, where $P(x) = -\frac{1}{x}$, $Q(x) = \cos(x)$, and $n=2$.

Divide the entire equation by $y^2$:

Let $v = y^{1-n} = y^{1-2} = y^{-1}$.
Differentiate $v$ with respect to $x$:

So, $y^{-2}\frac{dy}{dx} = -\frac{dv}{dx}$.
Substitute $v$ and $\frac{dv}{dx}$ into the transformed equation:

Multiply by $-1$ to get the standard linear form $\frac{dv}{dx} + P(x)v = Q(x)$:

For this linear equation, $P(x) = \frac{1}{x}$. The integrating factor $I(x)$ is:

(We assume $x > 0$ for simplicity).

Multiply $\frac{dv}{dx} + \frac{1}{x}v = -\cos(x)$ by $x$:

The left side is the derivative of the product $(xv)$:

Integrate both sides with respect to $x$:

To evaluate $\int x\cos(x) dx$, use integration by parts ($\int u\,dv = uv - \int v\,du$). Let $u=x$ and $dv=\cos(x)dx$. Then $du=dx$ and $v=\sin(x)$.

So,

Solve for $y$: