Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Find the particular solution \(y=f(x)\) that satisfies the differential equation and initial condition. $$ f^{\prime}(x)=\frac{1}{5} x-2 ; \quad f(10)=-10 $$

Step-by-Step Solution

Answer

The particular solution that satisfies the given differential equation and initial condition is \(f(x) = \frac{1}{10} x^2 - 2x -5\).

Step 1: Integral Calculation

The differential equation is given as \(f^{\prime}(x) = \frac{1}{5} x - 2\). To find \(f(x)\), we integrate \(f^{\prime}(x)\) with respect to \(x\). This gives \(f(x) = \frac{1}{10} x^2 - 2x + C\), where \(C\) is the constant of integration.

Step 2: Use the Initial Condition

Now we substitute the initial condition \(f(10)=-10\) into this equation. This gives \(-10 = \frac{1}{10} *(10)^2 - 2*10 + C\).

Step 3: Solve for C

Solving for \(C\) gives \(C = -5\).

Step 4: Write the Final Solution

Now, substitute \(-5\) for \(C\) in \(f(x) = \frac{1}{10} x^2 - 2x + C\). This gives the particular solution as \(f(x) = \frac{1}{10} x^2 - 2x -5\).

Key Concepts

Initial Value Problems

Initial value problems are a kind of differential equation that come with a condition that needs to be satisfied at a specific point. These problems are crucial in mathematics because they allow us to find a unique solution from many potential ones.

Imagine you're working with a differential equation like in the exercise above: \( f'(x) = \frac{1}{5}x - 2 \). This equation tells you how the rate of change of a function behaves. However, to pinpoint exactly which function you're dealing with, you require more information—like a snapshot of the function at a particular point.

Here's where the initial condition, \( f(10) = -10 \), comes into play. This condition allows us to determine a curve that precisely passes through the point \( (10,-10) \). Solving an initial value problem means determining the specific path (function \( f(x) \)) that not only abides by the rate of change given by the differential equation but also goes through the specified point.

Integration

Integration is the powerful mathematical tool we use to find functions from their rate of change. Given a differential equation, like \( f'(x) = \frac{1}{5}x - 2 \), integration allows us to "reverse" the differentiation process, retrieving \( f(x) \) from \( f'(x) \).

In this specific example, integrating \( f'(x) \) with respect to \( x \) leads us to the function: \( f(x) = \frac{1}{10}x^2 - 2x + C \). Notice the appearance of \( C \), the constant of integration. This constant springs from the indefinite nature of integration, which always leaves room for an unknown constant because when differentiating a constant, it vanishes to zero.

Thus, without additional conditions, there could be many functions that are solutions to the differential equation. When faced with such a problem, the power of integration complemented by initial conditions helps us pinpoint the exact function.

Particular Solution

A particular solution is the unique solution that exactly satisfies both the differential equation and the given initial condition. After you integrate a differential equation, you end up with a general solution, which includes the constant of integration \( C \).

To move from a general to a particular solution, you employ the initial condition specified in the initial value problem. For our case, using the condition \( f(10) = -10 \), we substitute \( x = 10 \) and solve \( -10 = \frac{1}{10}(10)^2 - 2(10) + C \).

Working through this, we find \( C = -5 \), leading to the particular solution \( f(x) = \frac{1}{10}x^2 - 2x - 5 \). This specific function satisfies not only the original differential equation but also the condition at the specified point. It's the tailored fit to the problem, a single path among many possibilities that precisely meets the defined criteria.