Problem 12 - Calculus

Question

Find the particular solution to the differential equation, given the general solution and an initial condition. ( \(C\) is the constant of integration.) $$P=C / t ; P=5 \text { when } t=3$$

Step-by-Step Solution

Answer

The particular solution is \( P = \frac{15}{t} \).

Step 1: Understand the General Solution

The general solution to the differential equation is given by \( P = \frac{C}{t} \). Here, \(P\) is a function of \(t\) and \(C\) is the constant of integration.

Step 2: Identify the Initial Condition

We are provided with the initial condition \( P = 5 \) when \( t = 3 \). This condition will help us find the specific value of the constant \(C\).

Step 3: Substitute the Initial Condition into the General Solution

Substitute \( P = 5 \) and \( t = 3 \) into the general solution \( P = \frac{C}{t} \):\[ 5 = \frac{C}{3} \]

Step 4: Solve for the Constant of Integration

Multiply both sides by 3 to solve for \(C\):\[ C = 5 \times 3 = 15 \]

Step 5: Write the Particular Solution

Substitute the value of \(C\) back into the general solution to find the particular solution:\[ P = \frac{15}{t} \]

Key Concepts

Initial Conditions

When solving differential equations, initial conditions are crucial for finding a specific solution tailored to a given scenario. These conditions are like a snapshot of your system at a specific point in time, often represented as a set of known values for dependent and independent variables. In our exercise, the initial condition is provided as \( P = 5 \) when \( t = 3 \). This particular point on the curve helps us determine the constant of integration \( C \), which differentiates the general solution from becoming overly generic. By substituting these values into our general equation, we restrain the infinite possibilities down to one exact solution. Initial conditions are essential in practical applications because they allow mathematical models to reflect real-world scenarios accurately.

Particular Solution

To obtain a particular solution of a differential equation, one must use the initial conditions to pinpoint a unique solution within the family of solutions described by the general solution. With general solutions often containing constants, particular solutions fix these constants to provide a specific answer that satisfies both the differential equation and its initial conditions. By substituting \( P = 5 \) and \( t = 3 \) into the equation \( P = \frac{C}{t} \), we calculate the particular value of the constant that meets these conditions, resulting in \( C = 15 \). Thereafter, the particular solution for this problem is \( P = \frac{15}{t} \). Particular solutions give invaluable insights into particular scenarios or behaviors of systems as they describe how a system behaves from a given starting point.

General Solution

In the realm of differential equations, general solutions are the most expansive form of a solution. They encompass an entire family of functions, characterized by one or more arbitrary constants such as \( C \). For linear differential equations, the general solution offers the broadest perspective of how solutions can vary before considering any specific initial conditions. In our example, the general solution is expressed as \( P = \frac{C}{t} \). Here, this equation can fit many specific scenarios, as \( C \) is a placeholder for any real number. Only when we apply initial conditions, such as \( P = 5 \) at \( t = 3 \), do we narrow down to a single particular solution. Hence, general solutions provide the theoretical groundwork from which any particular behaviors, influenced by initial conditions, can be drawn.