Differential Equation Exercise 12

Consider the following first - order differential equation:

$$ y' - x \cdot \cos x = 0 $$

This equation belongs to the class of separable differential equations of the form y′ + a(x)g(y) = 0, where \(a(x) = -x \cdot \cos x\) and \(g(y) = 1\).

We first write \(y′\) explicitly and switch to the standard derivative notation \(\frac{dy}{dx}\):

$$ y' = x \cdot \cos x $$

$$ \frac{dy}{dx} = x \cdot \cos x $$

Separating variables gives:

$$ dy = x \cdot \cos(x) \, dx $$

We now integrate both sides to determine \(y\):

$$ \int dy = \int x \cdot \cos(x) \, dx $$

The left-hand side integrates directly to \(y\):

$$ y = \int x \cdot \cos(x) \, dx $$

The right-hand side can be evaluated using integration by parts, \(\int f g' = fg - \int f' g\), choosing \(g' = \cos x\), so that \(g = \sin x\).

$$ y = x \cdot \sin x - \int 1 \cdot \sin x \, dx $$

$$ y = x \cdot \sin x - \int \sin x \, dx $$

Evaluating the remaining integral:

$$ y = x \cdot \sin x - (-\cos x) + c $$

$$ y = x \cdot \sin x + \cos x + c $$

We thus obtain the general solution of the differential equation:

$$ \boxed{y = x \cdot \sin x + \cos x + c} $$

This completes the solution.