$$ F^{\prime}(x)=\frac{d}{d x} \int_{a}^{x} g(t) d t=g(x) $$ to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval $$ I $$ of definition of the solution. $$ 2 x \frac{d y}{d x}-y=2 x \cos (x) ; \quad y=\sqrt{x} \int_{4}^{x} \frac{\cos (t)}{\sqrt{t}} d t $$ When $$ y=\sqrt{x} \int_{4}^{x} \frac{\cos (t)}{\sqrt{t}} d t $$, $$ \frac{d y}{d x}=\left(\frac{\cos (x)}{x} \times \frac{x \cos (t)}{\sqrt{t}} d t+\square\right. $$ Thus, in terms of $$ x $$, $$ 2 x \frac{d y}{d x}-y=(2 x \cos (x) $$

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To verify that the function
$$
y = \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt
$$
is a solution to the differential equation
$$
2x \frac{dy}{dx} - y = 2x \cos(x),
$$
we will follow these steps:

1. **Differentiate $$ y $$ with respect to $$ x $$ using the Product Rule.**
2. **Substitute $$ y $$ and $$ \frac{dy}{dx} $$ into the differential equation.**
3. **Simplify to show that both sides of the equation are equal.**

### Step 1: Differentiate $$ y $$ with Respect to $$ x $$

Given:
$$
y = \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt
$$

Apply the **Product Rule** for differentiation:
$$
\frac{dy}{dx} = \frac{d}{dx} \left( \sqrt{x} \right) \cdot \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt + \sqrt{x} \cdot \frac{d}{dx} \left( \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt \right)
$$

Compute each part separately:

1. **Differentiate $$ \sqrt{x} $$:**
$$
\frac{d}{dx} \left( \sqrt{x} \right) = \frac{1}{2\sqrt{x}}
$$

2. **Differentiate the integral using the **Fundamental Theorem of Calculus**:**
$$
\frac{d}{dx} \left( \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt \right) = \frac{\cos(x)}{\sqrt{x}}
$$

Now, substitute these results back into the derivative of $$ y $$:
$$
\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt + \sqrt{x} \cdot \frac{\cos(x)}{\sqrt{x}}
$$
$$
\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt + \cos(x)
$$

### Step 2: Substitute $$ y $$ and $$ \frac{dy}{dx} $$ into the Differential Equation

The differential equation is:
$$
2x \frac{dy}{dx} - y = 2x \cos(x)
$$

Substitute $$ y $$ and $$ \frac{dy}{dx} $$:
$$
2x \left( \frac{1}{2\sqrt{x}} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt + \cos(x) \right) - \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt = 2x \cos(x)
$$

### Step 3: Simplify the Equation

Distribute $$ 2x $$ in the left-hand side:
$$
2x \cdot \frac{1}{2\sqrt{x}} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt + 2x \cos(x) - \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt = 2x \cos(x)
$$
$$
x \cdot \frac{1}{\sqrt{x}} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt + 2x \cos(x) - \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt = 2x \cos(x)
$$
$$
\sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt + 2x \cos(x) - \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt = 2x \cos(x)
$$

Notice that the terms involving the integral cancel each other:
$$
\sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt - \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt = 0
$$
$$
0 + 2x \cos(x) = 2x \cos(x)
$$
$$
2x \cos(x) = 2x \cos(x)
$$

### Conclusion

After simplifying, both sides of the differential equation are equal:
$$
2x \cos(x) = 2x \cos(x)
$$
This confirms that the given function
$$
y = \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt
$$
is indeed a solution to the differential equation
$$
2x \frac{dy}{dx} - y = 2x \cos(x)
$$
on an appropriate interval $$ I $$ where the function is defined.
The function $$ y = \sqrt{x} \int_{4}^{x} \frac{\cos(t)}{\sqrt{t}} \, dt $$ is a solution to the differential equation $$ 2x \frac{dy}{dx} - y = 2x \cos(x) $$.

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