Question

Prove: If \(y_{p_{1}}\) is a particular solution of $$ P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x) $$ on \((a, b)\) and \(y_{p_{2}}\) is a particular solution of $$ P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{2}(x) $$ on \((a, b)\), then \(y_{p}=y_{p_{1}}+y_{p_{2}}\) is a solution of $$ P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x)+F_{2}(x) $$ on \((a, b)\)

Short answer

Question: Prove that if \(y_{p_1}\) is a solution to the differential equation \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x)\) and \(y_{p_2}\) is a solution to the differential equation \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{2}(x)\) on the interval \((a, b)\), then the sum \(y_p = y_{p_1} + y_{p_2}\) is a solution to the differential equation \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x) + F_{2}(x)\) on the same interval. Answer: We showed that by finding the derivatives of the summed solution and substituting them into the third differential equation, we could prove that \(y_p = y_{p_1} + y_{p_2}\) satisfies the equation \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x) + F_{2}(x)\) on the interval \((a, b)\).

Step-by-step solution

Write down the given information

We are given two particular solutions, \(y_{p_1}\) and \(y_{p_2}\), which are solutions to the following differential equations, respectively: $$ P_{0}(x) y_{p_1}^{\prime \prime}+P_{1}(x) y_{p_1}^{\prime}+P_{2}(x) y_{p_1}=F_{1}(x) \\ P_{0}(x) y_{p_2}^{\prime \prime}+P_{1}(x) y_{p_2}^{\prime}+P_{2}(x) y_{p_2}=F_{2}(x) $$ on the interval \((a, b)\).

Define the sum of the two particular solutions

Let \(y_p = y_{p_1} + y_{p_2}\). Our goal is to prove that \(y_p\) is a solution to: $$ P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x) + F_{2}(x) $$

Find derivatives of the summed solution

To show that \(y_p\) satisfies the third differential equation, we need to find its first and second derivatives. First derivative of \(y_p\): $$ y_p^{\prime} = y_{p_1}^{\prime} + y_{p_2}^{\prime} \\ $$ Second derivative of \(y_p\): $$ y_p^{\prime \prime} = y_{p_1}^{\prime \prime} + y_{p_2}^{\prime \prime} $$

Substitute the summed solution and its derivatives into the third differential equation

Now, we substitute \(y_p\), \(y_p^{\prime}\), and \(y_p^{\prime \prime}\) into the third differential equation: $$ P_{0}(x) (y_{p_1}^{\prime \prime} + y_{p_2}^{\prime \prime})+P_{1}(x)(y_{p_1}^{\prime} + y_{p_2}^{\prime})+P_{2}(x) (y_{p_1} + y_{p_2}) \\ = P_{0}(x) y_{p_1}^{\prime \prime}+P_{1}(x) y_{p_1}^{\prime}+P_{2}(x) y_{p_1} + P_{0}(x) y_{p_2}^{\prime \prime}+P_{1}(x) y_{p_2}^{\prime}+P_{2}(x) y_{p_2} \\ = F_{1}(x) + F_{2}(x) $$

Conclusion

Since the sum of the two particular solutions and their derivatives satisfies the third differential equation, we have proven that \(y_p = y_{p_1} + y_{p_2}\) is indeed a solution to: $$ P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x) + F_{2}(x) $$ on the interval \((a, b)\).

Key concepts

Particular Solution

When dealing with differential equations, a particular solution is one that satisfies the variable parts of the equation for a given set of functions. It doesn’t aim to solve the entire scope of the equation like a general solution, but rather fits perfectly under specific conditions or with specific inputs. Consider a differential equation of the form \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F(x)\).
This equation contains functions \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\), each multiplied by derivatives of \(y\) or \(y\) itself, set to equal \(F(x)\).

• A particular solution, \(y_p\), has been found when substituting it into the equation results in both sides equating.
• The particular solution helps satisfy one specific scenario or setting of \(F(x)\).
• Unlike the general solution, it won't contain arbitrary constants; it's precise and directly derived for the conditions given.

By exploring particular solutions, students learn how to understand specific cases of complex differential equations which aids in practical applications like engineering scenarios where specific outcomes are needed.

Superposition Principle

The superposition principle is a fundamental concept in mathematics, particularly in linear systems, including differential equations. It essentially describes that if two or more solutions of a linear system are identified, their sum is also a solution. Given our problem, if \(y_{p_{1}}\) is a solution to \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x)\), and \(y_{p_{2}}\) is a solution to \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{2}(x)\), then \(y_p = y_{p_1} + y_{p_2}\) is a solution to \(P_{0}(x) y^{\prime \prime}+P_{1}(x) y^{\prime}+P_{2}(x) y=F_{1}(x) + F_{2}(x)\).

• This principle is a direct result of linearity; superposition allows the understanding of complex scenarios by breaking them into simpler parts.
• It simplifies problems as you can work on parts, find their respective solutions, and combine them to form a complete picture.
• Students often use this principle to tackle bigger problems by first handling smaller, manageable pieces.

Understanding the superposition principle is key for students to grasp how linear systems work, making problem-solving more manageable and intuitive.

Linear Differential Equations

Linear differential equations are those in which the dependent variable and all its derivatives appear linearly. This means no powers or products of the dependent variable or its derivatives. The general form of a second-order linear differential equation is \(P_{0}(x) y^{\prime \prime} + P_{1}(x) y^{\prime} + P_{2}(x) y = F(x)\).

• The term "linear" refers to the fact that each term is either the constant, the dependent variable, or its derivative, each raised to the power of one.
• These equations are quite common in both theoretical mathematics and practical applications, such as physics and engineering.
• Solutions to these equations can be broken down into different components, like the particular and complementary solutions, which illustrate why understanding linear differential equations is foundational.

Linear differential equations offer a systematic approach to solving real-world problems by modeling them as linear systems, facilitating the application of mathematical techniques developed for such equations.