Question

Determining a Solution In Exercises \(21-28\) , determine whether the function is a solution of the differential equation \(x y^{\prime}-2 y=x^{3} e^{x} .\) $$ y=\cos x $$

Short answer

No, \(y = \cos x\) is not a solution to the differential equation \(xy' - 2y = x^3e^x\).

Step-by-step solution

Compute the derivative of the function

Given the function \(y = \cos x\), we use the differentiation formula to find its derivative (denoted by \(y'\)). The derivative of \(cos x\) is \(-sin x\), so we have \(y' = -sin x\).

Substitute into the differential equation

Next, we substitute the function \(y\) and its derivative \(y'\) into the differential equation \(xy' - 2y = x^3e^x\), giving us \(x*(-sin x) - 2(\cos x) = x^3e^x\). This simplifies to \(-xsin x - 2cos x = x^3e^x\).

Check if the equation holds

After simplifying, we can see that \(-xsin x - 2cos x = x^3e^x\) is not an identity. This means that, in general, the left-hand side is not equal to the right-hand side, implying that \(y = \cos x\) is not a solution to the differential equation \(xy' - 2y = x^3e^x\).

Key concepts

Derivative of Cosine

Understanding the derivative of cosine is essential when dealing with differential equations that involve trigonometric functions. In calculus, the derivative of a function at any point can be geometrically interpreted as the slope of the tangent line to the graph of the function at that point.

The function given in our exercise is the cosine function, denoted as \( y = \text{cos}(x) \). When we take the derivative of \( \text{cos}(x) \), denoted by \( \frac{d}{dx}(\text{cos}(x)) \) or simply \( y' \), we get \( y' = -\text{sin}(x) \). This rule comes directly from the standard derivatives in calculus, specifically the derivative of the cosine function.

Seeing the negative sign might be confusing at first, but it indicates that the sine curve, which represents the derivative, is out of phase with the cosine curve by \( \frac{\text{ }\pi}{2} \) radians and is decreasing wherever cosine is at its maximum or minimum. In the context of differential equations, this derivative allows us to relate the rates of change to the functions themselves, which is a cornerstone of differential calculus.

Substituting into Differential Equation

Once we have the derivative of our function, we can utilize it by substituting into the differential equation we're trying to solve. This step is pivotal as it integrates our specific function into the more general form of the equation we are given. In the exercise, we substitute \( y = \text{cos}(x) \) and its derivative \( y' = -\text{sin}(x) \) into the differential equation \( xy' - 2y = x^3e^x \).

Substitution can be straightforward, but care must be taken to ensure that all instances of the function and its derivatives are accurately replaced. The process involves replacing \( y \) and \( y' \) with their respective expressions and then simplifying. For our example, this means calculating \( x(-\text{sin}(x)) - 2(\text{cos}(x)) \), which simplifies to \( -x\text{sin}(x) - 2\text{cos}(x) \). This process transforms our differential equation into a form that we can analyze and compare to the given right-hand side of the original equation to check for validity.

Identifying Solutions to Differential Equations

Identifying solutions to differential equations is the process of verifying whether a given function satisfies the equation. After substituting the function and its derivative into the original equation, we reach a point where we either validate or invalidate the function as a solution.

In our exercise, after substitution and simplification, we compare both sides of the equation to see if they match. If the transformed equation is an identity, meaning it holds for all values of \( x \), then the function is a solution. However, as in our example, if the equation does not hold — since \( -x\text{sin}(x) - 2\text{cos}(x) \) is not equal to \( x^3e^x \) for all values of \( x \) — the function is not a solution. This comparison is crucial for confirming the suitability of prospective solutions to differential equations.

Differential Calculus

Differential calculus is the branch of mathematics that deals with the rate of change of quantities. It provides us tools like derivatives and differentials, which are vital for solving problems involving motion, growth, decay, and many more dynamic processes. The core concept of differential calculus is the derivative, which we can think of as a way to understand instantaneous rates of change, much like speed being the rate of change of distance with time.

Our example problem deeply relies on the principles of differential calculus. From finding the derivative of the cosine function to substituting it into a differential equation and checking for a solution, each step is rooted in differential calculus. Understanding these principles allows students to tackle a vast array of problems in physics, engineering, biology, and economics where such rates of change are fundamentally important.