Question

Verifying a Particular Solution In Exercises \(9-12\) , verify the particular solution of the differential equation. $$ \begin{array}{rl}{y=6 x-4 \sin x+1} & {y^{\prime}=6-4 \cos x} \\ {} & {y(0)=1}\end{array} $$

Short answer

After performing the computations, we established that the given function \(y=6x-4\sin x+1\) both has the correct derivative \(y'=6-4\cos x\), and meets the initial condition \(y(0)=1\), so it can be confirmed as a particular solution to the differential equation.

Step-by-step solution

Compute the derivative.

Compute the derivative of the given function \(y=6x-4\sin x+1\). Using the rules of differentiation, we obtain that the derivative of \(6x\) is \(6\), the derivative of \(-4\sin x\) is \(-4\cos x\), and the derivative of the constant \(1\) is \(0\). So, the derivative of the whole function is \(6-4\cos x\).

Compare computed derivative with given derivative.

The derivative we computed in step 1 is \(6-4\cos x\), which is exactly the derivative that was given in the problem statement \(y'=6-4\cos x\). So the first condition is validated and we can proceed with checking the second condition.

Check the initial condition.

We are given that the initial value of \(y\) when \(x=0\) is \(1\). Substitute \(x=0\) in the given function \(y=6x-4\sin x+1\) to compute \(y(0)\). Since \(\sin 0=0\), the function becomes \(y(0)=6*0-4*0+1=1\). So the second condition is also met.

Key concepts

Differential Equation

A differential equation is an equation that relates a function with its derivatives. In the context of our exercise, the function in question is expressed as \(y = 6x - 4\sin(x) + 1\). The purpose of the equation is to describe a relationship in which the rate of change (represented by the derivative) is tied to the function itself. Understanding differential equations is vital for modeling real-world phenomena where quantities change over time.

Differential equations come in various forms, such as ordinary differential equations (ODEs), where there is a single independent variable, and partial differential equations (PDEs), with two or more independent variables. Our particular problem involves an ODE, as it features one independent variable, 'x'. The verification process involves ensuring that the proposed solution meets the conditions set by the equation and any additional initial values, as we will see in further steps.

Derivative Computation

Derivative computation is a fundamental process in calculus, representing the rate at which a function changes at any given point. To verify a solution to a differential equation, we must show that the derivative of the purported solution matches the derivative presented in the equation.

In our specific problem, computing the derivative involves applying rules of differentiation to the function \(y\). This process includes differentiating terms like \(6x\) and \(\sin(x)\), leading us to the derivative \(y' = 6 - 4\cos(x)\). Derivative computation ensures that our solution is consistent with the behavior described by the differential equation, which is crucial for validation.

The complexity of derivative computation can vary, and advanced techniques such as the chain rule, product rule, and integration by parts may be required for more intricate functions. However, in our case, the straightforward application of basic differentiation rules suffices.

Initial Value Problem

An initial value problem is a specific type of differential equation where the solution is required to satisfy a given initial condition. The initial condition, in this case, could represent an initial state or starting point for the system described by the differential equation.

Our exercise presents us with the initial value \(y(0) = 1\), which is the value of our function when \(x = 0\). This initial condition is an integral part of the problem since it restricts the solution to not just any function that fits the differential equation, but one that specifically passes through the point \( (0, 1) \). Upon substituting \(x = 0\) into our function, we affirmed that the solution meets this condition, thereby satisfying not only the differential equation but also the prescribed initial state.

Initial value problems are common in physics and engineering, as they often represent practical situations where initial conditions are known and solutions must adhere to these constraints to be considered valid.