Question

If \(y(x)\) satisfies the differential equation \(y^{\prime}-y \tan x=2 x \sec x\) and \(y(0)=0\), then (A) \(y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}\) (B) \(y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}\) (C) \(y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9}\) (D) \(y^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}\)

Short answer

The correct answer is (A) \(y\left(\frac{\pi}{4}\right)=\frac{\pi^2}{8\sqrt{2}}\).

Step-by-step solution

Identify the Differential Equation

Recognize that the given differential equation is a first-order linear differential equation in the form \(y^\prime - y \tan x = 2x \sec x\).

Find the Integrating Factor

To solve a linear first-order differential equation, use an integrating factor \(\mu(x)\) which is given by \(\mu(x) = e^{\int P(x)\,dx}\), where \(P(x)\) is the coefficient of \(y\) in the differential equation. Here, \(P(x) = -\tan x\), so \(\mu(x) = e^{-\int \tan x\,dx} = e^{-\ln|\cos x|} = \frac{1}{\cos x} = \sec x\).

Multiply Throughout by the Integrating Factor

Multiply the entire differential equation by the integrating factor \(\sec x\) to obtain \(\sec x y^\prime - y \sec x \tan x = 2x\).

Recognize the Left Side as a Derivative

Observe that the left-hand side of the equation from Step 3 is the derivative of the product of the integrating factor \(\sec x\) and \(y\). Therefore, \(\sec x y^\prime - y \sec x \tan x = (\sec x y)^\prime\).

Integrate Both Sides of the Equation

Integrate both sides with respect to \(x\) to find \(\sec x y = \int 2x\,dx = x^2 + C\), where \(C\) is the constant of integration.

Solve for y

Divide by \(\sec x\) to solve for \(y(x)\): \((y \sec x)\prime = (x^2 + C)\) gives us \((y = x^2 \cos x + C \cos x)\).

Apply the Initial Condition

Use the initial condition \(y(0) = 0\) to find \(C\). Plugging these values into the solution gives us \(0 = 0 + C \cos(0)\), which implies that \(C = 0\). Therefore the solution is \(y = x^2 \cos x\).

Evaluate y at Specific Points

Now we need to evaluate \(y\) for the given points \(x =\frac{\pi}{4}\) and \(x=\frac{\pi}{3}\) to see which options are correct.

Check Option (A)

Evaluating \(y\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^2 \cos\left(\frac{\pi}{4}\right) = \frac{\pi^2}{16} \cdot \frac{\sqrt{2}}{2} = \frac{\pi^2}{16} \cdot \frac{8}{16 \sqrt{2}} = \frac{\pi^2}{8\sqrt{2}}\), which matches Option (A).

Differentiate y to Find y'

Differentiate \(y = x^2 \cos x\) with respect to \(x\) using the product rule to find \(y'\).

Evaluate y' at Specific Points

Specifically, find \(y'\left(\frac{\pi}{4}\right)\) and \(y'\left(\frac{\pi}{3}\right)\) to compare with Options (B) and (D).

Check Remaining Options

After computing \(y'\), plug in the specific values of \(x\) and match the results with the given options.

Key concepts

Integrating Factor Method

The integrating factor method is an essential tool for solving first-order linear differential equations. When a differential equation is given in the standard form \( y' + P(x)y = Q(x) \), the integrating factor \( \boldsymbol{\frac{1}{2}}\mu(x) \) will transform it into an exact equation that can be integrated directly. The integrating factor is defined as \( \boldsymbol{\frac{1}{2}}\mu(x) = e^{\boldsymbol{\frac{1}{2}}\int P(x)dx} \).To apply this method effectively:

• First, identify the function \( P(x) \) and calculate the integrating factor.
• Next, multiply the entire differential equation by \( \boldsymbol{\frac{1}{2}}\mu(x) \) to produce a product that can be recognized as the derivative of two functions.
• The left-hand side simplifies into a form that can be integrated with respect to \( x \).
• Finally, solve for \( y(x) \) and apply any initial conditions to determine the particular solution of the differential equation.

By using this process, complex differential equations can be solved systematically, providing clear steps towards the solution.

Initial Value Problem

An initial value problem (IVP) is a differential equation accompanied by a specified value, called the initial condition, which the solution must satisfy when \( x \) is at a particular point. In our exercise, the IVP is defined by the equation \( y^\boldsymbol{\frac{1}{2}} - y \tan x = 2x \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac('\), it is crucial to first determine the initial value of the function. In this case, the initial condition is \( y(0) = 0 \). Solving the IVP involves finding a function \( y(x) \) that satisfies both the differential equation and the initial condition. This often results in the introduction of a constant \( C \) after integration, which is then determined by plugging the initial condition into the general solution.

Differential Equation Solution

The solution to a differential equation represents a function or a family of functions that satisfy the equation. For first-order linear differential equations, the solution generally involves an arbitrary constant, which can be determined through initial conditions. In this context, after applying the integrating factor method, the differential equation is simplified to a point where it can be integrated to reveal the general solution.To form the complete solution, the process typically follows these steps:

• Integrating the modified equation after applying the integrating factor.
• Expressing the result in terms of \( y(x) \).
• Applying any given initial condition to resolve the constant of integration.

In the given exercise, the use of the initial condition \( y(0) = 0 \) led directly to the specific solution \( y = x^2 \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{1}{2}} \boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac{\boldsymbol{\frac{\frac{\frac('\). This solution satisfies both the equation and the initial value, proving it to be the correct solution.

Product Rule Differentiation

The product rule in differentiation is a crucial technique when dealing with products of functions. It states that if \( u(x) \) and \( v(x) \) are differentiable functions of \( x \), then the derivative of their product \( uv \) is given by \( \boldsymbol{\frac{1}{2}}\frac{d}{dx}(uv) = u\boldsymbol{\frac{1}{2}}\boldsymbol{\frac{1}{2}}\boldsymbol{\frac{1}{2}}\boldsymbol{\frac{1}{2}}\boldsymbol{\frac( v(x) + v\boldsymbol{\frac( u(x) \). The product rule is essential in finding the derivative of a product without having to expand the product first.For example, when differentiating the function \( y = x^2 \boldsymbol{\frac{1}{2}} \boldsymbol{\frac( x \), the product rule helps us find \( y' \) efficiently:

• Take \( u(x) = x^2 \) and \( v(x) = \boldsymbol{\frac( x \) as the two parts of the product.
• Apply the product rule to find \( y' = u'v + uv' \).
• Substitute the derivatives \( u' = 2x \) and \( v' = -\boldsymbol{\frac( x \) into the formula.

This produces the derivative of the function \( y \), which is needed to solve the differential equation or evaluate the function at specific points, as seen in the provided exercise.