Problem 51 - College Algebra and Calculus: An Applied Approach

Question

Find the particular solution \(y=f(x)\) that satisfies the differential equation and initial condition. $$ f^{\prime}(x)=2(x-1) ; \quad f(3)=2 $$

Step-by-Step Solution

Answer

The particular solution to the given differential equation and initial condition is \(f(x) = x^2 - 2x - 1\).

Step 1: Integrate the Differential Equation

To find \(f(x)\), we will integrate the given derivative with respect to \(x\). The integral of \(f^{\prime}(x) = 2(x-1)\) is \(f(x) = \int 2(x-1) dx = x^2 - 2x + C\), where \(C\) is the constant of integration.

Step 2: Apply the Initial Condition

Now, apply the given initial condition \(f(3) = 2\) to find the value of \(C\). Substituting these into the equation, we get: \(2 = 3^2 - 2*3 + C = 9 - 6 + C = 3 + C\). Solving for \(C\), we find \(C = -1\).

Step 3: Write down the Solution

Now we substitute \(C = -1\) into our equation to get the particular solution to the differential equation. As a result, \(f(x) = x^2 - 2x - 1\) is the solution that satisfies the given differential equation and initial condition.

Key Concepts

Integrating Differential Equations

Understanding the process of integrating differential equations is key to solving many problems in calculus. When faced with a differential equation such as \( f'(x) = 2(x-1) \), the goal is to find the function \( f(x) \) whose derivative is given by the equation. This process involves finding the antiderivative—or integral—of the function that is given.
Integrating is essentially the inverse operation to taking a derivative. If you know the derivative of a function, you can find the original function by integrating. In this example, the integral of \( 2(x-1) \) with respect to \( x \) is performed. The result is \( f(x) = x^2 - 2x + C \), where the \( +C \) represents the constant of integration. This constant is crucial as it encompasses all the infinite possible vertical shifts that the antiderivative may have.
To ensure the integration is done correctly, always remember to include the constant of integration. You can think of it as adding an unknown 'C' to catch any constant term that vanished in the differentiation process. Then, use initial conditions, if they are provided, to solve for the exact value of 'C' and turn the general solution into a particular solution.

Initial Conditions

Initial conditions allow you to pinpoint the exact solution when integrating a differential equation. In our example, the initial condition is the additional piece of information \( f(3) = 2 \). It specifies the value of the function at a particular point, acting as a key to unlock the value of the constant of integration.
When you have the general solution—remember that it includes the constant of integration 'C'—you apply the initial condition to find the specific 'C' that makes the solution unique to the problem at hand. This is effectively like saying, 'Out of all the possible functions that could work here, which one is ours?'
By substituting the initial condition into the general solution—here, replacing \( x \) with 3 and \( f(x) \) with 2—you employ the given information to find that \( C = -1 \). This gives you the particular solution that satisfies not only the differential equation but the provided initial condition as well.

Constant of Integration

The constant of integration 'C' in the integration process represents an indefinite number of possibilities. During integration, all the constants get absorbed into this symbol, and its value is not defined until we have more information. This is because when finding derivatives, any constant becomes zero, and this information is lost. The constant of integration accounts for that lost value.
In solving differential equations, after finding the indefinite integral, you use initial conditions to solve for 'C'. This transforms the general solution into a particular solution that is valid for a specific scenario. In our exercise, by using the initial condition \( f(3) = 2 \), we calculated that the constant 'C' must be -1 to satisfy the conditions given. Hence, integrating differential equations, applying initial conditions, and understanding the role of the constant of integration are all crucial steps in obtaining the full, precise solution to a differential equation.