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An explicit solution to a differential equation is a function expressed straightforwardly, without needing intricate methods or indirect steps. It is directly "spelled out," revealing the relationship between the variables even before solving the differential equation. For example, if your differential equation is given by \( y'' + y = 2 \cos x - 2 \sin x \), and you have a potential solution \( y = x \sin x + x \cos x \), showing that it satisfies the equation confirms it as an explicit solution.

Here’s how it works:

• Substitute the potential solution into the given differential equation.
• Simplify both sides of the equation.
• Ensure both sides balance or solve to the same expression.

Checking the explicit solution involves verifying every applicable term within the equation, as shown in the step-by-step solution where derivatives are found and substituted. If the given function remains true when plugged into the equation, it confirms it as an explicit solution.

The interval of definition in differential equations refers to the range of values over which the solution is valid and continuous. Like a map, it indicates where our solution can correctly be applied. For the function \( y = x \sin x + x \cos x \), composed of polynomial terms \( x \sin x \) and \( x \cos x \), its interval of definition usually considers the nature of the trigonometric and polynomial pieces.

Trigonometric functions, such as cosine and sine, are periodic and defined for all real numbers. When combined with polynomials, which also span all real values, the overarching solution \( y = x \sin x + x \cos x \) is viable for any real \( x \). This suggests that:

• The interval of definition for this solution is all real numbers \( (-\infty, +\infty) \).

Understanding intervals of definition ensures that your solutions to differential equations are used correctly, avoiding scenarios where solutions could become undefined or irrelevant.

The product rule is a principle of calculus used to find the derivative of two functions multiplied together. This concept is crucial in solving differential equations, particularly when dealing with functions that are products of simpler functions. The product rule tells us: if \( u(x) \) and \( v(x) \) are two differentiable functions, then the derivative of their product is:

• \((uv)' = u'v + uv' \)

In the context of the given problem, the function \( y = x \sin x + x \cos x \) can be tackled using the product rule to differentiate both components.
For the first term \( x \sin x \):

• \( u = x \) implies \( u' = 1 \)
• \( v = \sin x \) implies \( v' = \cos x \)

Thus, by applying the product rule, \((uv)' = 1 \cdot \sin x + x \cdot \cos x \).
For the second term \( x \cos x \):

• \( u = x \) implies \( u' = 1 \)
• \( v = \cos x \) implies \( v' = -\sin x \)

By applying the product rule again, \((uv)' = 1 \cdot \cos x + x \cdot (- \sin x) \).
By combining these results, as shown in our example, we arrive at the first derivative. Utilizing such fundamental calculus techniques properly is essential for solving more complex expressions and verifying solutions to differential equations.