Question

Solve the initial-value problems. $$ \frac{d y}{d x}=-8 x y^{2}+4 x(4 x+1) y-\left(8 x^{3}+4 x^{2}-1\right) ; \text { given solution } f(x)=x $$

Short answer

To verify that the given solution \(f(x)=x\) satisfies the initial-value problem, we calculate its derivative \(\frac{df(x)}{dx} = 1\). Then, we substitute \(f(x)\) and its derivative into the differential equation: \[ 1 = -8x^3 + 4x(4x^2 + x) - (8x^3 + 4x^2 - 1) \] Simplifying the equation, we obtain \(1 = 1\), confirming that the given solution is indeed a solution to the initial-value problem.

Step-by-step solution

Write down the differential equation and given solution

The first-order ordinary differential equation is given by: \[ \frac{dy}{dx} = -8xy^2 + 4x(4x+1)y - (8x^3 + 4x^2 - 1) \] The given function solution is: \[ f(x) = x \]

Calculate the derivative of the given solution

Let’s calculate the derivative \(\frac{df(x)}{dx}\): \[ \frac{df(x)}{dx} = \frac{d(x)}{dx} = 1 \]

Substitute the given solution and its derivative into the differential equation

Now we will substitute the given solution and its derivative into the original differential equation to verify that it is indeed a solution: \[ \frac{dy}{dx} = -8x(x)^2 + 4x(4x+1)(x) - (8x^3 + 4x^2 - 1) \]

Simplify the equation

Now, simplify and expand the equation: \[ 1 = -8x^3 + 4x(4x^2 + x) - (8x^3 + 4x^2 - 1) \] \[ 1 = -8x^3 + 16x^3 + 4x^2 - 8x^3 - 4x^2 + 1 \]

Check if the equation holds true

After simplification, we obtain: \[ 1 = 1 \] Since the equation holds true, the given solution \(f(x) = x\) is indeed a solution to the initial-value problem.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are a fundamental class of equations used in mathematics to describe the relationship between functions and their derivatives. An ODE involves ordinary derivatives of one or more dependent variables with respect to a single independent variable, often time or space.

In the context of our exercise, the ODE given is a first-order equation, meaning it involves the first derivative of the unknown function. The equation is non-linear because of the terms that include exponents and products of the unknown function, which is typical in many real-world problems ranging from physics to biology.

Understanding ODEs requires not just knowing how to solve them but also comprehending why they are modeled the way they are. They often encapsulate a physical law, like motion under gravity, population dynamics, or electrical circuits' behavior. With techniques such as separation of variables, integrating factors, and numerical methods, ODEs can be tackled effectively, allowing us to predict and interpret the dynamics of various systems.

Solution Verification

Solution verification is a crucial step in the process of solving a differential equation. After obtaining a potential solution, it’s important to confirm whether this solution actually satisfies the original problem, akin to checking your work in a standard math problem.

To verify a solution, one must substitute the proposed function back into the original differential equation. If, after simplification, the equation holds true for all points within the domain considered, the solution can be accepted as correct. This process ensures the integrity and accuracy of the solution found.

In our exercise, after substituting the solution function and its derivative back into the provided ODE, the verification yields a true statement. This signifies that the function proposed, in this case, a simple linear function, does indeed solve the differential equation for all values of the independent variable, providing a concrete demonstration of the solution’s validity.

Derivatives

Derivatives are one of the most powerful tools in calculus, representing the rate at which a function is changing at any given point. For instance, in physics, the derivative of position with respect to time gives an object's velocity, and the derivative of velocity with respect to time gives acceleration.

In this exercise, calculating the derivative of the function solution with respect to the independent variable is an essential step in the solution verification process. It allows us to express the change in the dependent variable in terms of the equation provided. Derivatives can also inform us about the function's behavior; for example, where its derivative is zero, the function has a critical point which may indicate a local maximum or minimum.

Understanding how to take derivatives correctly, not only in simple cases but also when dealing with products, quotients, or composite functions, allows us to solve a wide range of problems across various fields of study. With practice, the concept of derivatives becomes a fundamental part of one's mathematical toolkit.