Question

The solution of \(\mathrm{x}^{3}(\mathrm{dy} / \mathrm{dx})+4 \mathrm{x}^{2} \cdot \tan \mathrm{y}=\mathrm{e}^{\mathrm{x}} \cdot \mathrm{sec} \mathrm{y}\) satisfying \(\mathrm{y}(1)=0\) is: (A) \(\sin \mathrm{y}=\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1) \mathrm{x}^{-4}\) (B) \(\tan \mathrm{y}=(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}} \cdot \mathrm{x}^{-3}\) (C) \(\sin \mathrm{y}=\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1) \mathrm{x}^{-3}\) (D) \(\tan \mathrm{y}=(\mathrm{x}-2) \mathrm{e}^{\mathrm{x}} \cdot \log \mathrm{x}\)

Short answer

The derived solution for the given differential equation is \(\sin{y} = \sqrt{1 - \frac{x^8}{(e^x - e)^2}}\), which does not match any of the provided options. Therefore, the correct answer is not listed among the given options.

Step-by-step solution

Rewrite the given equation

We need to rewrite the given equation in a more familiar form for a first-order differential equation. Divide both sides of the equation by \(x^3\sec{y}\) to isolate the differential term: \[\frac{dy}{dx} + 4\frac{\tan y}{x} = \frac{e^x}{x^3}\]

Identify homogeneous equation

Observe that the equation obtained in step 1 is a first-order linear homogeneous differential equation as it can be rewritten in the form of: \[\frac{dy}{dx} + P(x)y = Q(x)\] In our case, \(P(x) = 4\frac{1}{x}\) and \(Q(x) = \frac{e^x}{x^3}\).

Calculate the integrating factor

The integrating factor (IF) is given by: \(IF = e^{\int P(x) dx}\) For our equation, this is: \(IF = e^{\int 4\frac{1}{x} dx}\) \(IF = e^{4\ln x}\) \(IF = x^4\)

Multiply the equation by the integrating factor

To solve the equation, we multiply both sides by the integrating factor: \(x^4\frac{dy}{dx} + 4x^3y = e^x\)

Integrate both sides of the equation

As the left side of the equation is an exact differential: \(\frac{d}{dx}(x^4y) = e^x\) Now, we integrate both sides of the equation with respect to x: \[\int (x^4y) dx = \int e^x dx\] \(x^4y = e^x + C\), where C is the integration constant.

Isolate y(x) and apply the initial condition

We need to isolate y(x) in the equation obtained in step 5: \(y(x) = \frac{e^x + C}{x^4}\) To find the value of C, let's apply the initial condition: \(y(1) = 0\): \(0 = \frac{e^1 + C}{1^4}\) So, \(C = -e\) The solution for y(x) is: \(y(x) = \frac{e^x - e}{x^4}\) Now, let's rewrite the solution in terms of sine or tangent as the answer choices are given in that format. We can recall the identity \(\sec{y} = \frac{1}{\cos{y}}\), and knowing that \(\sin^2{y} + \cos^2{y} = 1\), we can find \(\sin{y}\) or \(\tan{y}\). For our case: \[\frac{1}{\cos{y}} = \frac{e^x - e}{x^4}\] \[\cos^2{y} = \frac{x^8}{(e^x - e)^2}\] \[\sin^2{y} = 1 - \frac{x^8}{(e^x - e)^2}\] \[\sin{y} = \pm\sqrt{1 - \frac{x^8}{(e^x - e)^2}}\] Since \(y(1) = 0\), we have \(\sin y = 0\) which means \(\sin y\) has a positive value. \[\sin{y} = \sqrt{1 - \frac{x^8}{(e^x - e)^2}}\] This is as far as we can simplify y(x). Now let's compare our result with the given options. Upon comparison, we see that none of the provided options match the derived solution. As a result, we can conclude that the given options do not contain the correct answer.

Key concepts

First-order Linear Differential Equation

A first-order linear differential equation is the simplest type of differential equation to solve. Understanding its structure is crucial. Generally, it can be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Here, \(y\) is the dependent variable, and \(x\) is the independent variable. The functions \(P(x)\) and \(Q(x)\) depend only on \(x\). It is called 'first-order' because the highest derivative present is the first derivative, \( \frac{dy}{dx} \).

These equations are linear because they can be interpreted as straight lines in terms of \(y\) when separating \(x\), \(y\), and their derivatives. Solving them involves finding a function \(y\) that satisfies the equation for all values of \(x\).

This process typically requires calculating an integrating factor, enabling us to rewrite the differential equation to easily find the solution.

Integrating Factor

The method of solving first-order linear differential equations often involves finding an integrating factor. The integrating factor makes it possible to transform the equation into a more solvable form.

It is calculated from the function \(P(x)\) in the equation \( \frac{dy}{dx} + P(x)y = Q(x) \). Specifically, the integrating factor \(IF\) is determined by the formula \(IF = e^{\int P(x) dx}\). This special function, when multiplied through the entire differential equation, turns the left-hand side into an exact derivative. This transformation simplifies calculations significantly.

In the original exercise, we calculated the integrating factor as \(x^4\) by evaluating \( \int 4\frac{1}{x} dx\), leading to \(e^{4\ln x} = x^4\). Applying this integrating factor allowed us to manipulate the equation into an exact differential, making the integration process more straightforward.

Initial Conditions

Initial conditions are specific values provided for the solution of a differential equation that involve the dependent variable and potentially its derivatives at a particular point. They are crucial in finding the particular solution to a differential equation. Initial conditions usually look like \(y(x_0) = y_0\).

In our problem, the initial condition is \(y(1) = 0\). These conditions allow us to determine the constant of integration \(C\) that arises when integrating a differential equation. By substituting the initial values into the general solution, we can solve for \(C\), ensuring the solution matches the given initial state.

This adds precision to the solution, aligning it directly with the problem's context, and it often distinguishes a unique solution in scenarios where multiple solutions fit the general form of the differential equation.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables where both sides are defined. These identities are powerful tools in calculus and differential equations, providing ways to simplify and solve equations.

In trigonometry related to our differential equation solution, knowing that \( \sec(y) = \frac{1}{\cos(y)} \) helps. Using the identity \( \sin^2(y) + \cos^2(y) = 1 \), we can transform expressions and find equivalent forms in terms of \(\sin(y)\) or \(\tan(y)\).

For example, when solving for \(\sin(y)\) from \( \sec(y) \), we ended up with \( \sin(y) = \sqrt{1 - \frac{x^8}{(e^x - e)^2}} \). Such manipulations show how trigonometric identities effectively simplify expressions, and they play an instrumental role in finding compatible solutions to differential equations.