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In the context of differential equations, a particular solution is a specific solution that satisfies a given set of conditions, typically initial conditions provided in the problem statement. It stands out from the general solution, which contains arbitrary constants representing a family of functions. To find a particular solution, we substitute the initial conditions into the general solution to solve for these constants. In the given exercise, the particular solution is required for the differential equation: \[ y^{\prime} \sqrt{x} + \frac{1}{2} y = e^{\sqrt{x}} \] when \(x = 1\) and \(y = 3\). After finding the general solution, we apply these conditions to determine a specific value for the constant \(C\), giving us a particular solution that satisfies the initial conditions.

The integrating factor is a crucial tool for solving first-order linear ordinary differential equations. It is a function, typically denoted as \( \mu(x) \), that, when multiplied by the differential equation, transforms it into a form where the left side becomes the derivative of a product of functions. The integrating factor is determined by the equation:\[ \mu(x) = e^{\int P(x) \, dx} \]where \(P(x)\) is a coefficient from the standard linear form of the differential equation. For the exercise, the differential equation is rewritten as:\[ y^{\prime} + \frac{1}{2\sqrt{x}} y = \frac{e^{\sqrt{x}}}{\sqrt{x}} \]The integrating factor can then be calculated as \( e^{\sqrt{x}} \). Multiplying the entire differential equation by this factor, transforms it into a simpler expression, wherein the left-hand side is the derivative of \(e^{\sqrt{x}}y\), enabling straightforward integration.

Initial conditions are specific values provided in a differential equation problem that allow for the determination of a particular solution. They are typically given as values for the function and its derivatives at specific points. In this exercise, the initial conditions are:- \( x = 1 \)- \( y = 3 \)By utilizing these conditions, we can find the exact values of any unknown constants present in the general solution. Here, after solving the equation:\[ y = \frac{1}{2} e^{\sqrt{x}} + Ce^{-\sqrt{x}} \]we substitute \(x = 1\) and \(y = 3\) to determine the value of \(C\). The result gives us the specific form of the solution that satisfies the given initial conditions.

The standard form of a first-order linear differential equation is:\[ y^{\prime} + P(x)y = Q(x) \]Converting a given differential equation into this form is an essential step, as it simplifies the process of finding a solution. By identifying and separating the components \(P(x)\) and \(Q(x)\), one can apply techniques such as the integrating factor method. For the original equation in the exercise:\[ y^{\prime} \sqrt{x} + \frac{1}{2} y = e^{\sqrt{x}} \]we divide through by \(\sqrt{x}\), yielding the standard form:\[ y^{\prime} + \frac{1}{2\sqrt{x}} y = \frac{e^{\sqrt{x}}}{\sqrt{x}} \]This transformation facilitates the use of the integrating factor and helps systematically approach the solution of the differential equation.