Problem 11 - Differential Equations and Linear Algebra

Question

\((x-a)(x-b) y^{\prime}-(y-c)=0,\) where \(a, b, c\) are constants, with \(a \neq b.\)

Step-by-Step Solution

Answer

Rewrite the given differential equation in standard form: \[ y^{\prime} + \frac{y}{(x-a)(x-b)} = \frac{c}{(x-a)(x-b)}. \] Find an integrating factor: \[ \mu(x) = e^{\int \frac{1}{(x-a)(x-b)} dx}. \] Multiply the equation by the integrating factor and integrate both sides: \[ \int \mu(x) \cdot y'(x) dx + \int \mu(x) \cdot \frac{y}{(x-a)(x-b)} dx = \int \mu(x) \cdot \frac{c}{(x-a)(x-b)} dx. \] Define \(Y(x) = \mu(x) \cdot y(x)\) and solve for Y(x), then solve for y(x).

Step 1: Rewrite the given equation in standard form

The given equation needs to be rewritten in the standard form: \[ y' + P(x) y = Q(x). \] Dividing the given equation by \((x-a)(x-b)\) we get: \[ y^{\prime} - \frac{y-c}{(x-a)(x-b)} = 0. \] Now, we can rewrite the equation in the standard form as follows: \[ y^{\prime} + \frac{y}{(x-a)(x-b)} = \frac{c}{(x-a)(x-b)}. \]

Step 2: Find an integrating factor

An integrating factor is a function that helps transform a non-exact differential equation into an exact one, making it easier to solve. Let's find an integrating factor for the given equation. We have \(P(x) = \frac{1}{(x-a)(x-b)}\), and the integrating factor is given as the exponential of the integral of \(P(x)\): \(e^{\int P(x) dx}\). Calculating the integral and the exponential, we get the integrating factor: \[ \mu(x) = e^{\int \frac{1}{(x-a)(x-b)} dx}. \]

Step 3: Multiply the equation by the integrating factor

Now, we will multiply our equation by the integrating factor to get an exact differential equation: \[ e^{\int \frac{1}{(x-a)(x-b)} dx} \left( y^{\prime} + \frac{y}{(x-a)(x-b)} = \frac{c}{(x-a)(x-b)} \right). \]

Step 4: Integrate the resultant equation

Integrating both sides of the resultant equation, we obtain: \[ \int \mu(x) \cdot y'(x) dx + \int \mu(x) \cdot \frac{y}{(x-a)(x-b)} dx = \int \mu(x) \cdot \frac{c}{(x-a)(x-b)} dx. \]

Step 5: Solve for y

Now, we just need to rearrange the equation and solve for y: Let's define \(Y(x) = \mu(x) \cdot y(x)\), then we have: \[ Y'(x) \cdot (x - a)(x - b) + Y(x) = c \cdot \int \mu(x) \cdot \frac{1}{(x-a)(x-b)} dx, \] or, \[ Y'(x) \cdot (x - a)(x - b) + Y(x) = C(x), \] Now, we have an equation in terms of \(Y(x)\). We can solve for \(Y(x)\) and then further solve for the original variable, \(y(x)\). Please note that, due to the complexity of the integral involving the integrating factor, it may be necessary to use specialized integration techniques or numerical methods to fully solve for y(x).

Key Concepts

Integrating Factor

The integrating factor is a crucial function used to simplify solving linear differential equations. It transforms the original problem into an exact differential equation, making it easier to solve.

In our problem, we identified the linear differential equation with the standard form components, where \(P(x)\) is given by \(\frac{1}{(x-a)(x-b)}\).

• To find the integrating factor, we use the formula: \(\mu(x) = e^{\int P(x) dx}\).
• This involves computing the integral of \(P(x)\) and then taking the exponential function of the result.

The integrating factor essentially "balances" the equation, allowing us to rewrite it in a form that is integrable for both sides. This approach is significantly helpful because it reduces the complexity involved in dealing with non-exact equations, turning them into something more manageable.Once you multiply through the entire equation by the integrating factor, the left-hand side becomes an exact derivative, which can be integrated simply.

Linear Differential Equation

A linear differential equation is one of the most common types of differential equations and forms a foundation for understanding more complex systems. These equations are characterized by the linearity of their dependent variable and its derivatives.

In the initial step of our problem-solving process, we transformed a given equation into a linear differential equation by restructuring it to fit the standard form:\[y' + P(x)y = Q(x)\]

• This allowed us to identify \(P(x)\) and \(Q(x)\), critical components needed for finding the integrating factor.
• Linear equations of this form can always be solved using integrating factors, which aid in handling the non-homogeneous components.

Linear differential equations often appear in various fields, such as physics and engineering, where they describe systems that change at a rate proportional to their current state. Understanding how to work with them provides an essential tool in mathematical analysis and applied sciences.

Exact Differential Equation

An exact differential equation occurs when a differential equation can be represented in the form of a total derivative of some function. This is important because it allows us to integrate easily, finding an explicit solution for the dependent variable.

In the exercise given, after applying the integrating factor, our equation was transformed into an exact equation:

• The use of an integrating factor turned a non-exact equation into an exact one.
• This transformation made it possible to integrate both sides directly.

By multiplying the entire equation by the integrating factor, the individual components aligned more cleanly into an integrable format. As the equation becomes integrable, the solution can be found by evaluating the integral of the rearranged terms, yielding an expression for the function \(y(x)\). Exact differential equations are pivotal in calculus, as they extend the utility of integration to solve real-world problems.