Problem 7 - Single Variable Calculus

Question

In Exercises 7 and 8, the general solution of a differential equation is given. (a) Find the particular solution that satisfies the given initial condition. (b) Plot the solution curves correspond. ing to the given values of \(C\). Indicate the solution curve that corresponds to the solution found in part (a). $$ \begin{array}{l} x \frac{d y}{d x}+y=x^{3}, \quad y=\frac{C}{x}+\frac{x^{3}}{4} ; \quad y(1)=\frac{5}{4} \\ C=-2,-1,0,1,2 \end{array} $$

Step-by-Step Solution

Answer

To find the particular solution that satisfies the given initial condition y(1) = 5/4, we substitute x=1 and y=5/4 in the general solution equation: \( y = \frac{C}{x} + \frac{x^{3}}{4} \) Solving for C, we get: \( C = 1 \) So, the particular solution is: \( y = \frac{1}{x} + \frac{x^{3}}{4} \) To plot the solution curves for given values of C, consider: 1. For C = -2: \( y = \frac{-2}{x} + \frac{x^{3}}{4} \) 2. For C = -1: \( y = \frac{-1}{x} + \frac{x^{3}}{4} \) 3. For C = 0: \( y = \frac{0}{x} + \frac{x^{3}}{4} \) 4. For C = 1: \( y = \frac{1}{x} + \frac{x^{3}}{4} \) 5. For C = 2: \( y = \frac{2}{x} + \frac{x^{3}}{4} \) The solution curve that corresponds to the particular solution discovered in Step 1 is: \( y = \frac{1}{x} + \frac{x^{rz3}}{4} \)

Step 1: STEP 1: Find the particular solution

First, we need to find the particular solution that satisfies the given initial condition y(1) = 5/4. To do this, substitute x=1 and y=5/4 in the general solution equation: \( y = \frac{C}{x} + \frac{x^{3}}{4} \) \( \frac{5}{4} = \frac{C}{1} + \frac{1^{3}}{4} \)

Step 2: STEP 2: Solve for the constant C

Now we will solve for C by isolating it: \( C = \frac{5}{4} - \frac{1}{4} \) \( C = 1 \) So, the particular solution with the initial condition y(1) = 5/4 is: \( y = \frac{1}{x} + \frac{x^{3}}{4} \)

Step 3: STEP 3: Plot the solution curves for given values of C

To plot the solution curves corresponding to the given values of C = -2, -1, 0, 1, 2, we will plot the general solution equation for each value of constant C: 1. For C = -2: \( y = \frac{-2}{x} + \frac{x^{3}}{4} \) 2. For C = -1: \( y = \frac{-1}{x} + \frac{x^{3}}{4} \) 3. For C = 0: \( y = \frac{0}{x} + \frac{x^{3}}{4} \) 4. For C = 1: \( y = \frac{1}{x} + \frac{x^{3}}{4} \) 5. For C = 2: \( y = \frac{2}{x} + \frac{x^{3}}{4} \) Plot all five curves along with the particular solution curve, \( y = \frac{1}{x} + \frac{x^3}{4} \), found in Step 2.

Step 4: STEP 4: Indicate the solution curve that corresponds to the solution found in part (a)

Since the particular solution is found out to be when C = 1, the solution curve that corresponds to the particular solution discovered in Step 1 is: \( y = \frac{1}{x} + \frac{x^{3}}{4} \) This curve represents the solution of the given differential equation that satisfies the initial condition y(1) = 5/4.

Key Concepts

Particular Solution

When solving differential equations, a particular solution is a single method among the many that solves the equation while also satisfying a given initial condition. It's akin to finding a suit perfectly tailored to an individual, rather than any suit off the rack.

For instance, when given the general solution for the differential equation \( y = \frac{C}{x} + \frac{x^{3}}{4} \), where \(C\) denotes the arbitrary constant, we are presented with a set of potential solutions. However, to identify the particular solution, we must apply the initial condition, here \( y(1) = \frac{5}{4} \). By inserting \(x=1\) and \(y=\frac{5}{4}\) into the general solution and solving for \(C\), we find the value that tailors the general solution to fit this specific condition: \(C=1\). As a result, the particular solution becomes \( y = \frac{1}{x} + \frac{x^{3}}{4} \).

Initial Condition

The term initial condition refers to a specific criterion applied to a differential equation's solution, usually providing a value for the function or its derivatives at a certain point. Think of it as the 'starting line' in a race, determining where and how the solution begins.

In our example, the initial condition is \( y(1) = \frac{5}{4} \). When we input this condition into the general solution of the differential equation, it helps us to pinpoint the exact value of the constant \(C\), which leads us to the particular solution that starts from the prescribed 'starting line.' The initial condition essentially shapes the general solution to match the specific circumstances described by the condition.

Plotting Solution Curves

The process of plotting solution curves of differential equations involves visually representing the various solutions on a graph. This can illuminate the behavior of different solutions under variable conditions and the relationship between them.

To plot these curves, we graph the function for several values of the constant \(C\). Through this approach, we obtain a family of curves on the coordinate plane, each representing a potential solution to the differential equation. When plotting the curves for \(C = -2, -1, 0, 1,\) and \(2\) as mentioned in the steps above, we can observe how the solutions behave and how they change with varying \(C\) values. This visualization makes understanding the dynamic nature of differential equations much more intuitive.

General Solution of Differential Equation

The general solution of a differential equation represents the complete set of all possible solutions. It generally includes one or more arbitrary constants, which, when given specific values, can produce any particular solution. Like a chef's master recipe with room for variations, the general solution provides the framework from which all specific outcomes can be derived.

In our case, the general solution of the differential equation \( x \frac{dy}{dx} + y = x^{3} \) is \( y = \frac{C}{x} + \frac{x^{3}}{4} \). This equation includes the constant \(C\), which allows for an infinite number of solutions based on its value. By choosing different values of \(C\), we can explore how the general solution's shape and properties change, providing a powerful way to understand the general behavior of the system described by the differential equation.