Question

Find the solution of the following initial value problems. $$f^{\prime}(x)=2 x-3 ; f(0)=4$$

Short answer

Question: Find the solution to the initial value problem (IVP) involving a first-order ordinary differential equation (ODE) and an initial condition given by $f'(x) = 2x - 3$, $f(0) = 4$. Answer: The solution to the initial value problem is $f(x) = x^2 - 3x + 4$.

Step-by-step solution

Integrate the ODE

To find the general solution of the given ODE, we need to integrate both sides with respect to x: $$\int (f^{\prime}(x) \mathrm{d}x = \int (2x - 3) \mathrm{d}x$$ Evaluate both integrals: $$f(x) = x^2 - 3x + C$$ where C is the constant of integration.

Apply initial condition

Now we apply the initial condition, which states that \(f(0) = 4\). Plug in x=0 and f(x)=4 into the general solution equation: $$4 = 0^2 - 3(0) + C$$ Solving for C: $$C = 4$$

Final Solution

Insert the value of C back into our general solution to get the particular solution: $$f(x) = x^2 - 3x + 4$$ The solution to the initial value problem is \(f(x) = x^2 - 3x + 4\).

Key concepts

Ordinary Differential Equations

In calculus, ordinary differential equations (ODEs) involve functions and their derivatives. They are essential in modeling how something changes over time or space. ODEs appear in physics, engineering, economics, biology, and many other areas. In the equation \(f'(x) = 2x - 3\), we're dealing with a first-order differential equation, meaning it contains the first derivative of the unknown function \(f(x)\). Ordinary differential equations can describe simple systems like motion under constant acceleration or more complex phenomena like the growth of populations.

• The order of an ODE is determined by the highest derivative involved. For first-order ODEs, only the first derivative appears.
• The solution to an ODE can provide the function itself, revealing the behavior of a system over time.
• Initial conditions are crucial for solving ODEs, as they help find particular solutions specific to certain scenarios.

Understanding ODEs is fundamental in being able to analyze systems that change dynamically.

Integration

Integration is a core operation in calculus, used to find antiderivatives or areas under curves. In the context of an ordinary differential equation like \(f'(x) = 2x - 3\), integration is used to reverse the process of differentiation. By integrating the equation, we obtain the general solution \(f(x) = x^2 - 3x + C\). This solution reveals the antiderivative of the function's derivative.

• Indefinite integration leads to a family of functions, each differing by a constant \(C\).
• Definite integration, on the other hand, calculates the area under a curve between two given points, but is not used here.
• Finding the antiderivative helps in forcing us to trace back from rates of change to the original phenomena described.

Mastering integration is essential because it connects rates of change back to the quantities undergoing change, a key aspect in solving ODEs.

Constant of Integration

The constant of integration reflects the ambiguity inherent in antiderivatives: there are infinitely many functions that differentiate to the same result. Therefore, when we find the antiderivative and integrate, we always add \(C\) to our solution. This is the constant of integration. In our problem, after integrating, we obtain \(f(x) = x^2 - 3x + C\). To solve completely, we use the initial condition \(f(0) = 4\), plugging in \(x = 0\) to find \(C\).

• The constant of integration allows us to account for all possible vertical shifts of an antiderivative graph.
• Applying initial or boundary conditions enables us to determine a specific value for \(C\), thus fully solving the equation.
• Initial conditions anchor our solution to reality, providing a particular solution to the differential equation based on a known data point.

Understanding the constant of integration is crucial in differentiating between general and particular solutions.