Question

In Exercises \(16-24\) find the general solution. $$ y^{\prime}+\frac{4}{x-1} y=\frac{1}{(x-1)^{5}}+\frac{\sin x}{(x-1)^{4}} $$

Short answer

Question: Find the general solution of the given first-order linear differential equation: \(y'(x) + \frac{4}{x-1} y(x) = \frac{1}{(x-1)^5} + \frac{\sin x}{(x-1)^4}\). Answer: The general solution of the given differential equation is \(y(x) = \frac{1}{(x-1)^4} \left(\int\left((x-1)+(x-1)^4\sin x\right) \, dx + C\right)\), where \(C\) is the integration constant.

Step-by-step solution

Calculate the integrating factor

Find the integrating factor by computing \(e^{\int P(x) \, dx}\): $$ e^{\int \frac{4}{x-1} \, dx} = e^{4\int \frac{1}{x-1} \, dx} = e^{4 \ln (x-1)} = (x-1)^4 $$ So, the integrating factor is \((x-1)^4\).

Multiply by the integrating factor and integrate

Multiply both sides of the differential equation by the integrating factor, \((x-1)^4\): $$ (x-1)^4 y'(x) + 4(x-1)^3 y(x) = (x-1) + (x-1)^4 \sin x $$ Now, notice that the derivative of \(y(x) (x-1)^4\) is just the left-hand side of the modified equation. Therefore, we can write this as: $$ \frac{d}{dx}[y(x)(x-1)^4] = (x-1) + (x-1)^4 \sin x $$ Integrating both sides with respect to \(x\) yields: $$ y(x)(x-1)^4 = \int\left((x-1)+(x-1)^4\sin x\right) \, dx + C $$ where \(C\) is the integration constant.

Solve for y(x)

Finally, solve for \(y(x)\) by dividing both sides by \((x-1)^4\): $$ y(x) = \frac{1}{(x-1)^4} \left(\int\left((x-1)+(x-1)^4\sin x\right) \, dx + C\right) $$ This is the general solution of the given differential equation. Note that the integral inside the parentheses cannot be expressed in terms of elementary functions, so we leave it as it is.

Key concepts

Integrating Factor

The integrating factor is an essential tool for solving linear first-order ordinary differential equations (ODEs). It simplifies the equation, making it easier to solve. The core idea is to multiply the ODE by a specially chosen function. This function turns the left-hand side into a perfect derivative, which can then be integrated easily.

To find the integrating factor, you'll start with the equation in the form of \( y' + P(x)y = Q(x) \). The integrating factor \( \mu(x) \) is found using the expression:

• \( \mu(x) = e^{\int P(x) \; dx} \)

In our specific exercise, \( P(x) = \frac{4}{x-1} \). Calculating the integrating factor, we get \( \mu(x) = (x-1)^4 \). By multiplying the whole equation by this integrating factor, the left side becomes the derivative of \( y(x)(x-1)^4 \).

This simplifying feature of the integrating factor is what allows us to transform and solve the ODE efficiently.

General Solution

The general solution of a differential equation encapsulates all possible solutions by including a constant of integration. This constant allows for the inclusion of any initial condition possibly given at a later stage.

In our solved example, the equation becomes separable after applying the integrating factor. It now forms a complete derivative on the left-hand side. Thus, integrating both sides gives:

• \( y(x)(x-1)^4 = \int \left((x-1) + (x-1)^4 \sin x \right) \; dx + C \)

Here, \( C \) represents the constant of integration, indicating the generality of the solution. We eventually solve for \( y(x) \) by dividing through by \( (x-1)^4 \).

The result represents the general solution. It may seem complicated because the integral involved cannot be expressed using elementary functions, but it is complete in terms of finding a one-size-fits-all solution for all initial conditions.

Ordinary Differential Equations (ODEs)

Ordinary Differential Equations (ODEs) are fundamental in mathematics. They describe how a function's derivatives behave based on the function itself and a variable, often time or space.

There are various types of ODEs, but this specific exercise deals with a first-order linear ODE. These equations are of great importance in modeling real-world phenomena, as they can depict growth rates, decay, and many other dynamic behaviors in science and engineering.

In our exercise, we were given an ODE in the form: \( y' + P(x)y = Q(x) \). To solve it, we applied an integrating factor, transforming it into a form where the solution was more readily achieved by integrating.

Understanding the properties of ODEs, such as order and linearity, can significantly aid in knowing which method to apply for a solution. Mastering these concepts can greatly widen one's ability to handle various mathematical models and situations effectively.