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A first-order linear differential equation is an equation that involves a function and its first derivative. Understanding its structure is crucial for solving such problems. The general form of a first-order linear differential equation is given by:\[\frac{dy}{dx} + P(x)y = Q(x)\]Here,

• \(y\) is the dependent variable (usually a function of \(x\)).
• \(\frac{dy}{dx}\) represents the derivative of \(y\) with respect to \(x\).
• \(P(x)\) and \(Q(x)\) are functions of \(x\) that vary depending on the problem.

In this scenario, we begin by identifying the linear nature of the differential equation through rearranging the slope expression \(\frac{dy}{dx} = y + x\) to match this general format: \(\frac{dy}{dx} - y = x\). Recognizing this form enables us to systematically apply an integrating factor approach to find the solution. Don't worry if this seems complex at first; it becomes clearer with practice and understanding of each part's function.

To resolve a first-order linear differential equation, the integrating factor method is a powerful tool. This method turns a seemingly intricate equation into something more convenient to solve. Here's how it works step-by-step:

• Identify the standard form: Compare your equation to \(\frac{dy}{dx} + P(x)y = Q(x)\).
• Calculate the integrating factor, \(\mu(x)\), which is \(e^{\int P(x)dx}\).
• Multiply the entire differential equation by this integrating factor.

In our exercise, the standard form is \(\frac{dy}{dx} - y = x\). Here, \(P(x) = -1\), so the integrating factor is:\[\mu(x) = e^{-\int 1\, dx} = e^{-x}\]We then multiply each term in the equation by \(e^{-x}\):\[e^{-x}\frac{dy}{dx} - e^{-x}y = xe^{-x}\]This allows us to recognize the left side as the derivative of \(ye^{-x}\). Integration gives us:\[ye^{-x} = -xe^{-x} - e^{-x} + C\]Thus solving the initial equation by following these clearly laid out steps. Each piece eases the transition from complication to clarity.

An initial value problem (IVP) involves a differential equation paired with conditions that specify the value of the solution at a particular point. This is crucial for finding a unique solution.In this exercise, we are given that the curve passes through the point (0,1). This translates into needing to meet the condition \(y(0) = 1\). By substituting these values into the found general solution \(y = -x - 1 + Ce^{x}\), and solving for \(C\), we ensure the solution fits the initial conditions.Here's the simple calculation:

• Substitute \(x = 0\) and \(y = 1\) into the equation.
• Solve the resulting equation to find \(C\):\[1 = -0 - 1 + C \Rightarrow C = 2\]
• Introduce the \(C\) back into the equation, giving the unique solution:\[y = -x - 1 + 2e^{x}\]

By applying this approach, initial conditions guide us to the exact curve fitting the specified point and slope. Thus blending the abstract solution into a tangible path on the \(xy\)-plane.