Question

In Exercises \(30-37\) solve the initial value problem. $$ (x-1) y^{\prime}+3 y=\frac{1}{(x-1)^{3}}+\frac{\sin x}{(x-1)^{2}}, \quad y(0)=1 $$

Short answer

Answer: The solution to the initial value problem is \(y(x)=\frac{x-\cos x -2}{(x-1)^3}\).

Step-by-step solution

Identify the integrating factor

An integrating factor for a first-order linear differential equation is given by the function $$ \rho(x) = \exp\left(\int P(x) dx\right). $$ For our problem, we need to calculate the integrating factor using our given function \(P(x)=\dfrac{3}{x-1}\). We have $$ \rho(x) = \exp\left(\int \frac{3}{x-1} dx\right). $$

Calculate the integral of the integrating factor

We need to integrate the function \(\dfrac{3}{x-1}\) with respect to \(x\). We have $$ \mathrm{I}=\int \frac{3}{x-1} dx = 3\int \frac{1}{x-1} dx =3\ln(x-1) + C. $$ Since we only need the integral without the constant, we could write this as $$ \mathrm{I}=3\ln(x-1). $$

Determine the integrating factor

Using the function integration we calculated before, now we can find the integrating factor \(\rho(x)\), $$ \rho(x) = \exp\left(3\ln(x-1)\right). $$ Using the property \(\exp(\ln (u^n))=u^n\), for any \(u\) and \(n\), we have $$ \rho(x)=(x-1)^3. $$

Multiply the equation by the integrating factor

Now, we will multiply both sides of the equation by the integrating factor \((x-1)^3\): $$ (x-1)^3 (x-1) y' + 3(x-1)^3 y = 1+\sin x. $$

Integrate both sides of the adjusted equation

Our differential equation is now an exact equation (\((x-1)^3 y)'= 1+\sin x\). Integrate both sides with respect to \(x\): $$ \int (x-1)^3 y' dx + 3\int (x-1)^3 y dx = \int (1+\sin x) dx. $$ The integral of \((x-1)^3 y'\) is just \((x-1)^3 y\). Now we need to find the integral of the other terms. $$ \int (1+\sin x) dx = x - \cos x + C. $$ So, we have: $$ (x-1)^3 y = x-\cos x +C. $$

Solve for y(x)

Now, solve for the function y(x): $$ y(x)=\frac{1}{(x-1)^3}(x-\cos x+C). $$

Apply the initial condition

We are given the initial condition \(y(0)=1\). Plug in \(x=0\) into our function and solve for the constant \(C\): $$ 1=\frac{1-C}{(-1)^3}. $$ This gives \(C=-2\). Therefore, the final solution to the given initial value problem is: $$ y(x)=\frac{x-\cos x -2}{(x-1)^3}. $$

Key concepts

Integrating Factor

The concept of an integrating factor is a powerful tool for solving first-order linear differential equations. In essence, it's a function, often denoted as \(\rho(x)\), which we multiply by the original differential equation to transform it into an exact equation, which is easier to solve. The integrating factor is determined by the equation \(\rho(x) = \exp\left(\int P(x) \, dx\right)\), where \(P(x)\) is a function that only involves the independent variable, typically derived from the standard linear differential equation format \(y' + P(x)y = Q(x)\).

The actual calculation of the integrating factor requires integrating \(P(x)\), which in our exercise involves the integral of \(\frac{3}{x-1}\), resulting in \(\rho(x) = (x-1)^3\) after simplifying the exponentiation of the natural logarithm. Clearly understanding how to find the integrating factor is crucial since it simplifies the original equation, making the subsequent steps viable. It's a valuable strategy not just for textbook exercises but for various applications in mathematical modeling.

First-Order Linear Differential Equation

A first-order linear differential equation is characterized by its linearity, meaning it can be written in the form \(y' + P(x)y = Q(x)\) where \(y'\) denotes the derivative of \(y\) with respect to \(x\), and \(P(x)\) and \(Q(x)\) are functions of the independent variable \(x\). These equations are called 'first-order' because they involve only the first derivative of the unknown function.

When solving such an equation, the goal is to isolate and solve for the function \(y(x)\). The initial value provided, such as \(y(0)=1\) in our exercise, serves as a critical piece of information, allowing for the determination of any constants that arise during the integration process. These types of problems are incredibly common in fields like physics, engineering, and economics because they naturally describe many phenomena regarding rates of change, like velocity in mechanics or capital growth in economics.

Exact Equation

The term exact equation refers to a certain type of differential equation that is typically constructed by multiplying an inexact equation by an integrating factor. What makes an equation 'exact' is when it can be expressed in the form \(\mathrm{d}(M(x,y)) = 0\) where \(M(x,y)\) is a function whose differential matches the left-hand side of the original equation.

In the context of our exercise, after determining and applying the integrating factor, the resulting equation \(\mathrm{d}((x-1)^3 y)= 1+\sin x\) is exact because now the left side of the equation represents the derivative of a single entity. Integrating the adjusted equation often results in a more tangible expression, to which you can apply the initial conditions. Understanding the identification and use of exact equations greatly simplifies solving complex differential equations, and mastering this concept can immensely benefit students in their mathematical studies.