Question

Verifying a Particular Solution In Exercises \(9-12\) , verify the particular solution of the differential equation. $$ \begin{array}{rl}{y=\sin x \cos x-\cos ^{2} x} & {2 y+y^{\prime}=2 \sin (2 x)-1} \\ {} & {y\left(\frac{\pi}{4}\right)=0}\end{array} $$

Short answer

The function \(y=\sin x \cos x-\cos ^{2} x\) and its derivative \(y'=\cos^2x-\sin^2x+\sin2x\) satisfy the given differential equation and also confirm that \(y\left(\frac{\pi}{4}\right)=0\). This verifies the particular solution.

Step-by-step solution

Function Differentiation

We start by differentiating \(y=\sin x \cos x-\cos ^{2} x\) using the chain rule, obtaining \(y'=\cos^2x-\sin^2x+\sin2x\).

Substitute original and differentiated functions into the Differential equation

Next, substitute \(y\) and \(y'\) into the differential equation \(2y+y'=2\sin(2x)-1\) giving us \(2(\sin x \cos x-\cos ^{2} x) + (\cos^2x-\sin^2x+\sin2x) = 2\sin(2x)-1\). Simplifying this, we get: \(2\sin x\cos x-2\cos^2x+\cos^2x-\sin^2x+\sin2x = \sin2x-1\). Then we simplify the left hand side by canceling out equal terms and gathering like terms, resulting in \(\sin2x-1\), which is equal to the right hand side.

Verify at Given 'x' Value

Finally, we substitute the given value \(x=\frac{\pi}{4}\) into \(y=\sin x \cos x-\cos ^{2} x\). Substituting \(x=\frac{\pi}{4}\) into \(y\) we get 0 which validates that \(y\left(\frac{\pi}{4}\right)=0\).

Key concepts

Function Differentiation

Differentiation of a function is a fundamental concept in calculus used to determine how a function changes at any point. When we say we're differentiating a function, we're looking to find its derivative—the expression that gives us the rate of change. In this context, we started with the function \(y = \sin x \cos x - \cos^2 x\) and differentiated it to find its derivative \(y'\). Here's how we approach it:

• Break down the function: The function \(y = \sin x \cos x - \cos^2 x\) is composed of two parts: \(\sin x \cos x\) and \(-\cos^2 x\).
• Differentiate each part: We must differentiate each term separately. The derivative of \(\sin x \cos x\) is found using the product rule, and \(-\cos^2 x\) is differentiated using the chain rule.
• Combine the results: Bringing the differentiated parts together gives us \(y' = \cos^2 x - \sin^2 x + \sin 2x\).

Understanding function differentiation is crucial because it is often the first step in solving differential equations effectively. By breaking down functions into manageable components, we apply rules like the product and chain rules to find derivatives easily.

Chain Rule

The chain rule is a vital differentiation technique used when a function is composed of one or more functions. This rule helps to find the derivative of a composite function. Here is a simple way to understand it:Consider the case of the function \(-\cos^2 x\).

• Identify the outer and inner functions: Here, the outer function is \(u^2\) where \(u = \cos x\).
• Differentiate the outer function: The derivative of \(u^2\) is \(2u\).
• Differentiate the inner function: The derivative of \(\cos x\) is \(-\sin x\).
• Combine using the chain rule: Multiply these derivatives: \(2u \times (-\sin x) = -2 \cos x \sin x\).

The chain rule is essential in calculus as it simplifies the differentiation of complex functions into more manageable steps. In our function, this rule allows us to differentiate \(-\cos^2 x\), showing its importance in solving our differential equation.

Particular Solution

A particular solution in differential equations is a solution that satisfies the differential equation and fits specific initial conditions or constraints. In this exercise, the particular solution of the given differential equation is \(y = \sin x \cos x - \cos^2 x\), and it must satisfy the condition \(y\left(\frac{\pi}{4}\right) = 0\).Here's how we verify it:

• Substitute into the differential equation: We use the function and its derivative in the equation \(2y + y' = 2\sin(2x) - 1\).
• Simplify and compare: After simplification, the expressions on both sides of the equation must be equal, confirming that the function satisfies the differential equation.
• Check initial condition: Substitute \(x = \frac{\pi}{4}\) into the function \(y\). It should result in the value 0, which matches the given condition.

The process of verifying a particular solution ensures that we find not just any solution to a differential equation, but the specific one that aligns with additional information provided in the problem.

Sin and Cos Functions

The sine and cosine functions are fundamental trigonometric functions pivotal in the study of waves, oscillations, and circles. In calculus, these functions are crucial components due to their periodic nature and differentiability.Let's look at their roles in differentiation:

• Basic derivatives: The derivative of \(\sin x\) is \(\cos x\), and the derivative of \(\cos x\) is \(-\sin x\).
• Combination in equations: Functions like \(\sin x \cos x\) arise frequently and can be differentiated using product or chain rules.
• Employment in identities: Trigonometric identities, like \(\sin 2x = 2\sin x \cos x\), often simplify differentiation and integration processes.

Understanding the properties and derivatives of sine and cosine functions is invaluable in calculus, as it allows for the simplification of complex problems. These functions' periodic properties are particularly useful in verifying and solving differential equations, as seen in this exercise.