Question

According to Theorem \(2.1 .2,\) the general solution of the linear nonhomogeneous equation $$ y^{\prime}+p(x) y=f(x) $$ is $$ y=y_{1}(x)\left(c+\int f(x) / y_{1}(x) d x\right). $$ where \(y_{1}\) is any nontrivial solution of the complementary equation \(y^{\prime}+p(x) y=0 .\) In this exercise we obtain this conclusion in a different way. You may find it instructive to apply the method suggested here to solve some of the exercises in Section 2.1 . (a) Rewrite (A) as $$ [p(x) y-f(x)] d x+d y=0, $$ and show that \(\mu=\pm e^{\int p(x) d x}\) is an integrating factor for (C). (b) Multiply (A) through by \(\mu=\pm e^{\int p(x) d x}\) and verify that the resulting equation can be rewritten as $$ (\mu(x) y)^{\prime}=\mu(x) f(x) $$ Then integrate both sides of this equation and solve for \(y\) to show that the general solution of (A) is $$ y=\frac{1}{\mu(x)}\left(c+\int f(x) \mu(x) d x\right). $$ Why is this form of the general solution equivalent to (B)?

Short answer

In summary, to find the general solution of the given linear nonhomogeneous equation, we can follow these steps: 1. Rewrite the equation as: $[p(x) y - f(x)] dx + dy = 0$ 2. Calculate the integrating factor, $\mu(x)$, which is determined by: $\mu = \pm e^{\int p(x) dx}$ 3. Multiply equation (A) by the integrating factor. 4. Verify that the resulting equation can be rewritten as: $(\mu(x) y)^{\prime} = \mu(x) f(x)$ 5. Integrate both sides of the equation: $\int (\mu(x) y)^{\prime} dx = \int \mu(x) f(x) dx$ 6. Solve for \(y\): $y = \frac{1}{\mu(x)} \left(c + \int f(x) \mu(x) dx\right)$ This approach yields the same general solution as the one provided in the question, demonstrating that the methods are equivalent.

Step-by-step solution

(a) Finding the integrating factor

To find the integrating factor for the rewritten equation \((A)\), let's first rewrite it as: $$ [p(x) y - f(x)] dx + dy = 0 $$ Now for a given equation of the form \(M dx + N dy = 0\), an integrating factor \(\mu\) is such that \(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = \frac{N}{\mu}\) or \(\frac{M}{\mu}\). Here, \(M = p(x) y - f(x)\), and \(N = 1\). Taking the partial derivatives, we get \(\frac{\partial M}{\partial y} = p(x)\) and \(\frac{\partial N}{\partial x} = 0\). This implies that: $$ \frac{N}{\mu} = p(x) \; or \; \frac{M}{\mu}=-p(x). $$ From the above relation, we can calculate the integrating factor \(\mu\): $$ \mu = \pm e^{\int p(x) dx} $$

(b) Solving for y using the integrating factor

Multiplying equation (A) by the integrating factor \(\mu(x)\): $$ \pm e^{\int p(x) dx} [p(x) y - f(x)] dx \pm e^{\int p(x) dx} dy = 0 $$ Let's now verify that the resulting equation can be rewritten as: $$ (\mu(x) y)^{\prime} = \mu(x) f(x) $$ Differentiating the left side with respect to x, we get: $$ (\pm e^{\int p(x) dx} y)^\prime = \pm e^{\int p(x) dx} y^\prime \pm p(x) e^{\int p(x) dx} y = \pm e^{\int p(x) dx} [p(x) y - f(x)] + \pm e^{\int p(x) dx} y^\prime $$ Since the equation we obtained on differentiating the left side is equal to the equation after multiplying equation \((A)\) by the integrating factor, thus the equation can be rewritten as: $$ (\mu(x) y)^{\prime} = \mu(x) f(x) $$ Now, integrating both sides: $$ \int (\mu(x) y)^{\prime} dx = \int \mu(x) f(x) dx $$ Which gives: $$ \mu(x) y = c + \int f(x) \mu(x) dx $$ Now, we can solve for \(y\): $$ y = \frac{1}{\mu(x)} \left(c + \int f(x) \mu(x) dx\right) $$ The given form of the general solution is equivalent to the one provided in the question, as both of them express \(y\) in terms of a nontrivial solution of the complementary equation, plus an integral term involving \(f(x)\). This shows that the methods are essentially equivalent and lead to the same general solution for the given linear nonhomogeneous equation.

Key concepts

Integrating Factor

An integral component in solving linear differential equations is finding the integrating factor. This factor, denoted as \( \mu \), is a special function introduced to turn a non-exact differential equation into an exact one. In simpler terms, it acts as a multiplier to balance the equation, making it solvable by integrating. The integrating factor is derived from the equation \( y' + p(x)y = f(x) \) by calculating \( \mu = e^{\int p(x) dx} \). This expression ensures that, when multiplied through the original differential equation, the left-hand side becomes the derivative of \( \mu(x) y \), thus simplifying the solving process.

Essentially, the integrating factor transforms the differential equation so that it can be easily integrated, hence solving for \( y \). When applied correctly, it simplifies the work required to find the general solution by making the equation integrable in a straightforward manner.

Nonhomogeneous Equation

A nonhomogeneous differential equation is one where the function on the right side of the equation is not zero. In mathematical terms, it takes the form \( y' + p(x)y = f(x) \), where \( f(x) \) is a non-zero function. This inclusion of \( f(x) \) differentiates it from homogeneous equations, where \( f(x) \) would be zero.

In tackling a nonhomogeneous equation, the goal is to find a particular solution that satisfies the equation. This is achieved by working with methods such as the integrating factor or the method of undetermined coefficients, which leverage the structure of \( f(x) \).

The nonhomogeneous term \( f(x) \) usually represents external forces or inputs in practical applications, making these equations highly relevant in real-world scenarios. Understanding how to approach these equations is crucial for deriving solutions that reflect the true behavior of the system involved.

Complementary Equation

The complementary equation is a crucial part of solving a nonhomogeneous differential equation. It is derived from setting the nonhomogeneous term \( f(x) \) to zero, resulting in the homogeneous equation \( y' + p(x)y = 0 \).

Solving this complementary equation helps in finding the general solution \( y_1(x) \), which forms the basis for solving the entire nonhomogeneous equation. Typically, the solutions to the complementary equation involve exponential functions that can capture the natural response of the system without external input.

This solution, \( y_1(x) \), is essential as it represents the self-driven behavior of the system, or how it naturally evolves. In many cases, once the complementary solution is found, it acts as a pivot point for integrating the particular solution associated with the nonhomogeneous part of the original equation, thereby composing the general solution.