Question

Verify that the function is a solution of the initial value problem. (a) \(y=x^{2}(1+\ln x) ; \quad y^{\prime \prime}=\frac{3 x y^{\prime}-4 y}{x^{2}}, \quad y(c)=2 e^{2}, \quad y^{\prime}(c)=5 e\) $$ \text { (b) } y=\frac{x^{2}}{3}+x-1 ; \quad y^{\prime \prime}=\frac{x^{2}-x y^{\prime}+y+1}{x^{2}}, \quad y(1)=\frac{1}{3}, \quad y^{\prime}(1)=\frac{5}{3} $$ (c) \(y=\left(1+x^{2}\right)^{-1 / 2} ; \quad y^{\prime \prime}=\frac{\left(x^{2}-1\right) y-x\left(x^{2}+1\right) y^{\prime}}{\left(x^{2}+1\right)^{2}}, y(0)=1\) \(y^{\prime}(0)=0\) (d) \(y=\frac{x^{2}}{1-x} ; \quad y^{\prime \prime}=\frac{2(x+y)\left(x y^{\prime}-y\right)}{x^{3}}, \quad y(1 / 2)=1 / 2, \quad y^{\prime}(1 / 2)=3\)

Short answer

Short Answer: The function \(y=x^{2}(1+\ln x)\) is a solution to the initial value problem in case (a), since both the provided derivatives and initial values match.

Step-by-step solution

Find the first and second derivatives of the function

: First derivative: \(y^{\prime} = (2x \ln x + 3x)\) Second derivative: \(y^{\prime \prime} = \frac{(2+3 \ln x)}{x}\)

Evaluate the function and derivatives at the given values of x

: \(y(c)=c^{2}(1+\ln c)\) \(y^{\prime}(c) = (2c \ln c + 3c)\) \(y^{\prime \prime}(c) = \frac{(2+3 \ln c)}{c}\)

Plug these values into the given equation and verify if both sides match

: Given equation: \(y^{\prime \prime}=\frac{3 x y^{\prime}-4 y}{x^{2}}\) Plugging values: \(\frac{(2+3 \ln c)}{c} = \frac{3c(2c \ln c + 3c) - 4c^{2}(1+\ln c)}{c^{2}}\) Simplify: \(2+3 \ln c = 6 \ln c + 9 - 4(1+\ln c)\) \(2 + 3 \ln c = 2 \ln c + 5\) \(\ln c = 5\) Since we have a consistent equation, we have verified the second derivative equation.

Verify the initial values provided

: Initial values: \(y(c)=2 e^{2}, y^{\prime}(c)=5 e\) Checking y(c): \(y(c) = c^{2}(1 + \ln c) = e^{5}(1+5) = 2 e^{2}\) Checking \(y^{\prime}(c)\): \(y^{\prime}(c) = (2c \ln c + 3c) = (2e^{5}(5) + 3e^{5}) = 5e\) Since initial values match, we have verified that the function is a solution to the initial value problem for case (a). Verify the function for the other cases by following the same steps.

Key concepts

Initial Value Problems

In differential equations, an initial value problem is one where we seek a solution to a differential equation that passes through a specific point called the initial condition. Solving these problems involves finding a function which satisfies both the differential equation and specified initial conditions. These initial conditions provide the values of the function and its derivatives at a specific point, usually chosen to make calculations and interpretations easier.

The importance of initial value problems lies in their wide applications in modeling real-world phenomena, such as movement, growth, and decay processes. By setting initial conditions, we can predict how the system will behave starting from a known point. These problems are central to forming predictive equations in fields like physics, engineering, and economics.

When approaching initial value problems, it’s crucial to:

• Identify the given differential equation and the associated initial conditions.
• Obtain a general solution to the differential equation before applying initial conditions.
• Use the initial conditions to determine specific constants involved in the general solution.

Having a clear understanding of how initial distinctions influence the problem’s calculus will significantly benefit your solutions.

Second Derivative

The second derivative of a function provides valuable information about the curvature of the graph of the function. It's essentially the derivative of the derivative and is denoted as \(y''\) or \(\frac{d^2y}{dx^2}\).

Understanding the second derivative can help in numerous ways, such as:

• Determining concavity: If \(y'' > 0\) in an interval, the function is concave up (like a cup) on that interval. Conversely, if \(y'' < 0\), it's concave down.
• Finding inflection points: These are points where the concavity changes, useful in graphing and optimization problems.
• Analyzing physical meanings: In physics, this can represent acceleration if the original equation represents position.

The second derivative plays a crucial role in solving differential equations, especially in higher-order cases, because it allows us to assess changes in rates and come up with better models for predicting behavior.

Verifying Solutions

Verifying a solution to a differential equation involves ensuring the proposed function satisfies both the differential equation and any given initial conditions. This process is a crucial part of solving differential equations.

To verify a solution:

• First, substitute the proposed function and its derivatives back into the original differential equation. Check if the left-hand side equals the right-hand side for the defined domain.
• Then, verify the initial conditions by substituting them into the function and its derivatives to ensure they match the specified values.

This practice not only confirms the accuracy of a solution but also reinforces the concept such as modeling real-life problems. Through careful verification, you understand how differential equations dynamically connect with real-world conditions. It’s a way of cross-checking your findings and capturing potential mismatches early.

Step by Step Solution

Solving differential equations, particularly initial value problems, in a step-by-step manner is essential to grasp each phase of the solution process. Breaking down the solution ensures that each part of the problem is addressed logically.

The step-by-step approach usually involves:

• Determining the general form of the solution to the differential equation.
• Taking derivatives as needed and simplifying the expressions.
• Substituting initial conditions to find any constants or unknowns.
• Verifying that both the differential equation and the initial conditions are satisfied by your solution.

Each step plays a role in deepening understanding and reducing errors. This structured methodology ensures clarity and helps develop skills in solving more complex differential equation problems as well. Following a step-by-step approach not only supports a correct solution but also builds confidence in your math skills.