Problem 13 - Calculus

Question

Find the particular solution to the differential equation, given the general solution and an initial condition. ( \(C\) is the constant of integration.) \(y=\sqrt{2 t+C} ;\) the solution curve passes through (1,3)

Step-by-Step Solution

Answer

The particular solution is \( y = \sqrt{2t + 7} \).

Step 1: Identify the Initial Condition

The given initial condition is that the solution curve passes through the point (1, 3). This means that when \( t = 1 \), \( y = 3 \). We will use this information to find the particular value of \( C \).

Step 2: Substitute Initial Condition into the General Solution

Substitute \( t = 1 \) and \( y = 3 \) into the general solution, \( y = \sqrt{2t + C} \). Doing this gives \( 3 = \sqrt{2(1) + C} \).

Step 3: Solve for the Constant C

Square both sides of the equation obtained from Step 2 to eliminate the square root: \( 3^2 = 2 \cdot 1 + C \). This simplifies to \( 9 = 2 + C \). Subtract 2 from both sides to find \( C = 7 \).

Step 4: Write Out the Particular Solution

Now that we have determined \( C = 7 \), substitute this value back into the general solution to get the particular solution: \( y = \sqrt{2t + 7} \).

Key Concepts

Initial Condition

In the world of differential equations, an **initial condition** is an additional piece of information provided to help us find a specific solution from a family of possible ones. Imagine a family of curves that can be described by a differential equation. Without additional information, any one of these curves could be a solution. This is where initial conditions come in to help pinpoint the exact curve that should be considered.

For example, in our exercise, the given initial condition is a point that the solution curve passes through, specifically (1, 3). This means when the variable \( t \) equals 1, the variable \( y \) should equal 3. Initial conditions are crucial as they refine the general solution by providing a specific criterion that the solution must satisfy. They serve as a guiding factor to determine the constant of integration, \( C \), in our general solution. This guiding nature makes them indispensable in finding particular solutions.

General Solution

The **general solution** of a differential equation encompasses all possible solutions that satisfy the equation. It is usually expressed in terms of one or more arbitrary constants. These constants, like \( C \) in our exercise, represent a family of solutions giving rise to numerous curves.

In our specific exercise, the general solution is given by the function \( y = \sqrt{2t + C} \). This formula reflects all the potential solution curves corresponding to different values of the constant \( C \). The general solution provides a global understanding of the behavior of the equation and sets the stage for specific solutions.

• The form of the solution is dependent on the equation's structure.
• It contains arbitrary constants that are yet to be determined.

Determining the general solution lays the groundwork for deriving particular solutions, which fulfill additional conditions or constraints.

Particular Solution

A **particular solution** is derived from the general solution by applying initial conditions to find specific constant values, thus narrowing down from a set of possible solutions to one single solution tailored to a particular scenario. After determining the constant using the initial condition, the general solution adjusts to become this particular solution.

In the exercise, after plugging in the initial condition point (1, 3) into the general solution \( y = \sqrt{2t + C} \), we solved for the constant \( C \). Through this process, we found \( C = 7 \), resulting in the particular solution \( y = \sqrt{2t + 7} \).

This particular solution uniquely outlines the path of the original curve through the given point, representing the specific scenario indicated by the initial condition. With it, we can confidently describe the behavior of the differential equation for that specific case.