Question

Solve the initial value problem and find the interval of validity of the solution. $$ (x+1)(x-2) y^{\prime}+y=0, \quad y(1)=-3 $$

Short answer

Answer: The particular solution to the given initial value problem is \(y(x) = \frac{-6}{(x-2)(x+1)}\). The interval of validity is \(x \in (-\infty, -1) \cup (-1, 2) \cup (2, \infty)\).

Step-by-step solution

Identify the differential equation type and its components

The given differential equation is of the form \((x+1)(x-2)y'+y=0\). We can rewrite it as \(y'+\frac{1}{(x+1)(x-2)}y = 0\). This is a first-order linear differential equation with P(x) = \(\frac{1}{(x+1)(x-2)}\) and Q(x) = 0.

Find the integrating factor

To find the integrating factor, we need to calculate \(e^{\int P(x) dx} = e^{\int \frac{1}{(x+1)(x-2)} dx}\). This integral can be solved using partial fractions. Let's rewrite the integral as \(\int \frac{A}{x+1}+\frac{B}{x-2} dx\). Applying partial fractions, we obtain \(1 = A(x-2)+B(x+1)\). Solving for A and B, we get A = -1 and B = 1. Therefore, the integral becomes \(\int -\frac{1}{x+1}+\frac{1}{x-2} dx\) and its antiderivative is \(-ln|x+1|+ln|x-2|+C\). Taking the exponent of both sides, we get the integrating factor as: \(e^{\int P(x) dx} = \frac{|x-2|}{|x+1|}\). We can ignore the absolute values since it will be factored out after integration.

Apply the integrating factor to the differential equation

Multiply the entire differential equation by the integrating factor: \(\frac{x-2}{x+1}\left(y'+\frac{1}{(x+1)(x-2)}y\right) = 0\). This simplifies to \(y'(x-2)+y(x+1)=0\).

Integrate to find the general solution

Integrate both sides: \(\int (y'(x-2)+y(x+1)) dx = \int 0 dx\). This leads to the general solution: \(y(x-2)(x+1) = C\), where C is the constant of integration.

Apply the initial condition to find the particular solution

Given y(1) = -3, plug in x = 1 and y = -3 into the general solution: \(-3(1-2)(1+1) = C\). This gives C = -6. Therefore, the particular solution is \(y(x) = \frac{-6}{(x-2)(x+1)}\).

Find the interval of validity

The particular solution is valid as long as the denominator \((x-2)(x+1)\) is non-zero. Thus, the interval of validity is \(x \in (-\infty, -1) \cup (-1, 2) \cup (2, \infty)\). In conclusion, the solution to the initial value problem is \(y(x) = \frac{-6}{(x-2)(x+1)}\), and it's valid on the interval \(x \in (-\infty, -1) \cup (-1, 2) \cup (2, \infty)\).

Key concepts

First-Order Linear Differential Equations

First-order linear differential equations form the foundation for a wide range of problems in mathematics and engineering, due to their straightforward structure and solvability. A standard form for these equations is given as:
\[ y' + P(x)y = Q(x) \]
Here, \( y' \) denotes the derivative of \( y \) concerning \( x \), while \( P(x) \) and \( Q(x) \) are functions of \( x \). The solution to such an equation involves finding a function \( y(x) \) that satisfies the given relationship for all \( x \) within a certain interval.
When addressing this kind of differential equation, like in the provided exercise, our main goal is to isolate \( y \) and its derivative \( y' \) on one side, coming up with an equation that can be managed through known methods like integrating factors (which we will discuss shortly). The 'linearity' aspect implies that the equation doesn’t involve terms like \( y^2 \) or \( sin(y) \), making the equation more amenable to direct analytical solutions.

Integrating Factors

An integrating factor is a mathematical tool that transforms a non-exact first-order linear differential equation into an exact one by multiplying every term by a specially chosen function. The primary purpose of the integrating factor, \( \text{IF} \), is to enable us to rewrite our differential equation as the derivative of a product of functions, which can then be integrated much more straightforwardly.
In general, if we have a differential equation in the form \( y' + P(x)y = Q(x) \), the integrating factor is defined as:
\[ \text{IF} = e^{\int P(x) dx} \]
When applied correctly, the integrating factor turns the left side of our differential equation into the derivative of \( \text{IF} \times y \), thereby simplifying the process of integration. As shown in our exercise, after determining the integrating factor, we multiply it across the entire differential equation, which leads to a form that is ready for direct integration to solve for \( y \).

Partial Fractions

The method of partial fractions is an essential algebraic technique used to break down complex rational expressions into simpler fractions that are easier to integrate. This is especially useful when dealing with integrating factors that present as rational functions.
In the context of our exercise, to calculate the integrating factor, we had to integrate a fraction of the form \( \frac{1}{(x+1)(x-2)} \). To make this integration manageable, we expressed it as a sum of two simpler fractions:
\[ \frac{A}{x+1} + \frac{B}{x-2} \]
By determining the values of constants \( A \) and \( B \), we simplified the integral and found the integrating factor. Mastery of partial fractions can thus be invaluable, as it frequently serves as a bridge between a complex integrand and its antiderivative.

Interval of Validity

The interval of validity is critical to understanding the domain over which a solution to a differential equation is applicable. A solution may not be valid over the entire real number line due to points where the solution becomes undefined, such as divisions by zero or taking logarithms of negative numbers.
In our example, after finding the particular solution \( y(x) = \frac{-6}{(x-2)(x+1)} \), we have to ensure that the denominator is non-zero to avoid discontinuities or undefined expressions. Consequently, the values \( x = 2 \) and \( x = -1 \) cause the denominator to be zero and are therefore excluded from the interval of validity. As a result, we observe that the solution is valid over the intervals \( (-fty, -1) (x-2)(x+1) (-1, 2) (x-2)(x+1) (2, nfty) \). It's vital for students to consider these intervals to fully understand the range over which the solution to an initial value problem holds true.