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The integrating factor method is a powerful technique to solve first-order linear differential equations, which are equations of the form \(\frac{dy}{dx} + p(x)y = q(x)\). In these cases, the integrating factor, denoted by \(\mu(x)\), is an exponential function of the integral of \(p(x)\). Specifically, \(\mu(x) = e^{\int p(x)dx}\).

To apply the integrating factor method, you multiply every term of the differential equation by \(\mu(x)\), which allows you to express the left side of the equation as the derivative of the product of \(\mu(x)\) and \(y\). This simplifies the equation, making it easier to solve since you can then integrate both sides with respect to \(x\). Finally, you isolate \(y\) to get the solution to the original differential equation. The utility of the integrating factor lies in its ability to turn a difficult-to-solve differential equation into a form where direct integration is possible.

Integration by parts is a technique based on the product rule for differentiation and is used when you are dealing with the integral of the product of two functions, say \(u\) and \(dv\). The formula for integration by parts is \(\int u dv = uv - \int v du\).

The method involves choosing \(u\) and \(dv\) intelligently such that \(du\) and \(v\) (after integrating \(dv\)) lead to an integral that is simpler than the original. This method is particularly useful when you encounter an integral that is the product of a polynomial and an exponential function or trigonometric functions within the integrating factor method. For the equation in the exercise, after applying the integrating factor, you are left with integrals that can be solved by employing integration by parts, further facilitating the finding of a solution to the differential equation.

The substitution method in differential equations involves introducing a new variable, say \(v\), to replace an existing expression involving the original variables, often with the goal of simplifying the equation. This method is helpful when you spot patterns or relationships between the variables that can be exploited.

For example, if \(y\) is raised to a power or a function of \(y\) is multiplied by \(\frac{dy}{dx}\), you might choose \(v = y^2\) or a similar substitution. After substituting, the key is to express every part of the original differential equation in terms of the new variable and \(\frac{dv}{dx}\), simplifying to a form that is more manageable. When the new equation is solved for \(v\), you must remember to substitute back to the original variable to obtain the final solution to the problem. This method is especially beneficial when the original equation does not easily fit into other standard forms for differential equations.