91Ó°ÊÓ

A differential equation is an equation that relates a function with its derivatives. In our given problem, we have the equation \( \frac{dy}{dx} = \sqrt{5x+1} \). This tells us how the function \( y \) changes with respect to \( x \).

Differential equations are essential because they can describe various real-world phenomena, such as population growth, heat transfer, and physical motions. The ultimate goal is to find a function \( y(x) \) that satisfies the equation throughout a specific domain.

• **Ordinary Differential Equations (ODEs):** Involve one independent variable and its derivatives, like in our problem.
• **Partial Differential Equations (PDEs):** Involve multiple independent variables. These are more complex.

Integrating a differential equation means finding the original function from its derivative. The integration process helps us transform the differential equation into an algebraic equation that we can solve.

Integration by substitution is a powerful technique used to simplify integrals. In our exercise, we face the integral \( \int \sqrt{5x+1} \, dx \). Direct integration isn't straightforward, which is why we apply substitution.

To use this method, we substitute a part of the integral with a new variable. We let \( u = 5x + 1 \), making our integral easier to handle. Thus, \( du = 5 \, dx \) or \( dx = \frac{du}{5} \).

• **Step 1:** Identify the substitution: It often involves turning a complex expression into a simpler one.
• **Step 2:** Substitute and change the differential: Replace every \( x \) term with the new variable \( u \).
• **Step 3:** Integrate the simpler expression: This is usually more manageable now.

After substitution, the integral \( \int \sqrt{u} \frac{du}{5} \) becomes much more approachable, allowing us to solve it and eventually back-substitute to return to the original variable.

Initial conditions are crucial for solving differential equations because they help determine a unique solution. These conditions give specific values of the function at a certain point, like \( y(3) = -2 \) in our problem.

Without initial conditions, a differential equation generally has infinite solutions. This is because integrating introduces an arbitrary constant \( C \). Initial conditions help eliminate this ambiguity by "fixing" this constant to a specific value.

• **Purpose of Initial Conditions:** Ensure the solution of the differential equation fits given specific data points.
• **Finding Constant \( C \):** Substitute the given conditions into the general solution to solve for \( C \).

In our exercise, after integrating the differential equation and finding a general solution, we use the initial condition \( y(3) = -2 \) to solve for the constant \( C \). This step is essential in expressing the solution that meets the condition precisely and gives us a specific equation instead of a family of solutions.