Question

Assume that all functions in this exercise are defined on a common interval \((a, b)\). (a) Prove: If \(y_{1}\) and \(y_{2}\) are solutions of $$ y^{\prime}+p(x) y=f_{1}(x) $$ and $$ y^{\prime}+p(x) y=f_{2}(x) $$ respectively, and \(c_{1}\) and \(c_{2}\) are constants, then \(y=c_{1} y_{1}+c_{2} y_{2}\) is a solution of $$ y^{\prime}+p(x) y=c_{1} f_{1}(x)+c_{2} f_{2}(x) $$ (This is theprinciple of superposition.) (b) Use (a) to show that if \(y_{1}\) and \(y_{2}\) are solutions of the nonhomogeneous equation $$ y^{\prime}+p(x) y=f(x) $$ then \(y_{1}-y_{2}\) is a solution of the homogeneous equation $$ y^{\prime}+p(x) y=0 $$ (c) Use (a) to show that if \(y_{1}\) is a solution of \((\mathrm{A})\) and \(y_{2}\) is a solution of \((\mathrm{B}),\) then \(y_{1}+y_{2}\) is a solution of \((\mathrm{A})\)

Short answer

Question: Prove the principle of superposition in the given context of linear ordinary differential equations (ODEs) and use this principle to find the solutions of some other ODEs. Answer: Our solution demonstrated the principle of superposition by showing that a linear combination of two solutions of different ODEs is also a solution to another ODE. We also showed that the difference between two solutions of a nonhomogeneous equation is a solution of the homogeneous equation. Finally, we proved that the sum of solutions of two different ODEs is a solution of the first ODE.

Step-by-step solution

Part (a): Prove the principle of superposition

Given two ODEs with solutions \(y_1\) and \(y_2\): $$ y^{\prime}+p(x) y=f_{1}(x) \quad \text{and} \quad y^{\prime}+p(x) y=f_{2}(x) $$ We want to show that a linear combination of these two solutions, \(y=c_{1} y_{1}+c_{2} y_{2}\), is a solution of another ODE: $$ y^{\prime}+p(x) y=c_{1} f_{1}(x)+c_{2} f_{2}(x) $$ To prove this, differentiate \(y\) with respect to \(x\): $$ y^{\prime}=c_{1} y_{1}^{\prime}+c_{2} y_{2}^{\prime} $$ Now substitute this into the given ODE and simplify: $$ \begin{aligned} (c_{1} y_{1}^{\prime}+c_{2} y_{2}^{\prime})+p(x)(c_{1} y_{1}+c_{2} y_{2}) &= c_{1} (y_{1}^{\prime}+p(x)y_{1})+c_{2}(y_{2}^{\prime}+p(x)y_{2}) \\ &= c_{1}f_{1}(x)+c_{2}f_{2}(x) \end{aligned} $$ The equation holds true, proving the principle of superposition.

Part (b): Difference between two solutions of a nonhomogeneous equation

Let \(y_1\) and \(y_2\) be solutions of the nonhomogeneous equation: $$ y^{\prime}+p(x) y=f(x) $$ We want to show that \(y=y_1-y_2\) is a solution of the homogeneous equation: $$ y^{\prime}+p(x) y=0 $$ Using the superposition principle from part (a), let \(c_1 =1\) and \(c_2 = -1\). Then, we have: $$ y^{\prime}+p(x) y=f_1(x)-f_2(x) $$ Substitute back into the given nonhomogeneous equation and simplify: $$ y^{\prime}+p(x) y = f(x) - f(x) = 0 $$ Thus, \(y=y_1-y_2\) is a solution of the homogeneous equation.

Part (c): Sum of solutions of two different ODEs

Given two ODEs with solutions \(y_1\) and \(y_2\): $$ y^{\prime}+p(x) y=f_1(x) \quad \text{and} \quad y^{\prime}+p(x) y=f_2(x) $$ Using the superposition principle from part (a), let \(c_1 =1\) and \(c_2 = 1\). Then, we have: $$ y = y_1 + y_2 $$ We want to show that this sum is a solution of the first ODE. Substitute this into the first ODE and simplify: $$ \begin{aligned} (y_1+y_2)'+ p(x)(y_1+y_2) &= y_1'+y_2'+ p(x)y_1+p(x)y_2 \\ &=(y_1'+p(x)y_1)+(y_2'+p(x)y_2) \\ &=f_1(x)+f_2(x) \end{aligned} $$ Thus, the sum of solutions of two different ODEs, \(y_1+y_2\), is a solution of the first ODE.

Key concepts

Linear Differential Equations

In the realm of calculus, a linear differential equation is a crucial concept that represents a relationship involving a function and its derivatives. When we talk about linear differential equations, we are referring to equations that contain a function, say y(x), and its derivatives, which are constructed in such a way that if you were to draw them out, the function and its derivatives would not be multiplied together or taken to any power other than the first. In simpler terms, they play nicely and don't mix in a complicated way.

Think of a linear differential equation in the form of A(x)y'' + B(x)y' + C(x)y = D(x), where A(x), B(x), C(x), and D(x) could be any functions of x themselves, but they don't mess around with y or its derivatives. They might transform or stretch the equation, but they don't entangle it in a nonlinear dance. A special case to note is when D(x) equals zero – then the equation is called homogeneous; otherwise, it's nonhomogeneous.

Superposition Principle

Just like a DJ combines different tracks to create a new mix, the superposition principle allows us to mash up solutions of linear differential equations to find a brand new solution. This principle is a game-changer in the world of differential equations because it tells us that if you have two or more solutions to a linear equation, you can combine them (typically through addition or subtraction, and scaling by constants) to cook up another valid solution.

Imagine you have two functions, think of them as distinct musical notes, and these functions are solutions to two separate differential equations with similar structures. According to the superposition principle, if you were to compose these notes together—by maybe adding them up and scaling—we'd end up with a harmonious tune that's also a solution to a third differential equation, one that's related to the originals. This principle holds true as long as we're working with linear differential equations and it doesn't matter if they are homogeneous or nonhomogeneous – the principle still applies.

Homogeneous and Nonhomogeneous Equations

Differential equations are like two sides of a coin: on one side you have homogeneous equations, and on the flip side are the nonhomogeneous bunch. A homogeneous differential equation is kind of like a pure note in music, it's self-contained and doesn’t rely on an external force. In mathematical lingo, this means the equation equals zero when you remove all traces of the function and its derivatives. The homogeneous equation looks somewhat like y'' + p(x)y' + q(x)y = 0, where the right side is solely zero, denoting the absence of external forces or inputs.

On the contrary, a nonhomogeneous differential equation resembles a note being influenced by external effects. There's an additional function on the equation's stage that doesn’t involve the original function y or its derivatives, sort of like an added beat or rhythm. It has the form y'' + p(x)y' + q(x)y = g(x), where g(x) represents the external influence. The presence of g(x) makes the equation nonhomogeneous – it’s not purely about the function y by itself anymore, but it’s affected by something else too.

Solution of Differential Equations

Solving a differential equation can be a bit like untying a complex knot – it's all about isolating the function you're looking for and its derivatives. The solution process involves finding a function, or a set of functions, that satisfies the equation when you plug them into it. This solution can take various forms, like a specific function that works perfectly for your equation or, in the case of homogeneous linear differential equations, a whole family of functions that all work just as well.

These solutions are not just hypotheticals – they can describe real-world phenomena, such as the cooling of coffee over time or the movement of a pendulum. When we solve these equations, we're often trying to predict how something will behave, and the solutions give us the ability to do that. Depending on the complexity of the equation, we might be able to write down an exact solution, or we might have to approximate using computational methods. Either way, finding the solution to a differential equation is the key to unlocking the secrets that the equation holds about how things change and interact in the physical world.