Question

Find the general solution, given that \(y_{1}\) satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation.

Short answer

In this problem, we are given that \(y_1\) is a solution to the complementary equation of a linear second-order homogeneous differential equation. Our goal is to find the general solution for this complementary equation. The general solution is given by \(y(x) = (K_2 + K_3(Cx + K_1))y_1\), and a fundamental set of solutions is \(\{y_1, y_2\} = \{y_1, (Cx+K_1)y_1\}\), where \(K_2\) and \(K_3\) are arbitrary constants, and \(y_1\) and \(y_2\) are linearly independent solutions to the complementary equation.

Step-by-step solution

Identify Given Information

Given that \(y_1\) is a solution of the complementary equation.

Use \(y_1\) to Find another Linearly Independent Solution

We're looking for a linearly independent solution \(y_2\). We can use the method of reduction of order to find it. Given the equation: $$y_2 = v(x)y_1.$$ Differentiate \(y_2\) with respect to \(x\): $$y_2' = v(x)y_1' + v'(x)y_1.$$ Differentiate \(y_2'\) with respect to \(x\): $$y_2'' = v(x)y_1'' + 2v'(x)y_1' + v''(x)y_1.$$

Find the Complementary Equation

As \(y_1\) and \(y_2\) satisfy the complementary equation, we substitute their values into the homogeneous linear ODE, which takes the form: $$y'' + p(x) y' + q(x) y = 0.$$ Plug in values for \(y_2 = v(x)y_1\), \(y_2'=v(x)y_1' + v'(x)y_1\), and \(y_2''=v(x)y_1'' + 2v'(x)y_1' + v''(x)y_1\): $$[v(x)y_1'' + 2v'(x)y_1' + v''(x)y_1] + p(x)[v(x)y_1' + v'(x)y_1] + q(x)[v(x)y_1] = 0.$$

Eliminate \(v''\) Term and Solve for \(v'(x)\)

Observe that the \(v''(x)y_1\) term can be eliminated, which simplifies the equation to: $$v(x)y_1'' + 2v'(x)y_1' + p(x)v(x)y_1' + p(x)v'(x)y_1 + q(x)v(x)y_1 = 0.$$ Since \(y_1\) is a solution, it follows that: $$y_1''+ p(x)y_1'+q(x)y_1 = 0.$$ Consequently, the equation becomes: $$2v'(x)y_1' + p(x)v'(x) y_1 = 0.$$ Now, divide both sides by \(2y_1y_1'\): $$v'(x) \left(\frac{2y_1' + p(x)y_1}{2y_1 y_1'}\right) = 0.$$ It follows that \(v'(x)\) is a constant, say \(C\): $$v'(x) = C.$$

Integrate and Find \(v(x)\)

Integrate \(v'(x) = C\) with respect to \(x\): $$v(x) = Cx + K_1.$$ Here, \(K_1\) is the constant of integration.

Find the Second Linearly Independent Solution \(y_2\)

Now that we have found \(v(x)\), we can find the second linearly independent solution, \(y_2\): $$y_2 = v(x)y_1 = (Cx+K_1) y_1.$$

Construct the General Solution of the Complementary Equation

Finally, we construct the general solution as a linear combination of the two linearly independent solutions, \(y_1\) and \(y_2\): $$y(x) = K_2 y_1 + K_3 y_2 = K_2 y_1 + K_3 (Cx+K_1) y_1$$ Where \(K_2\) and \(K_3\) are arbitrary constants. So, the general solution is given by: $$y(x) = (K_2 + K_3(Cx + K_1))y_1.$$ And a fundamental set of solutions for the complementary equation is \(\{y_1, y_2\} = \{y_1, (Cx+K_1)y_1\}\).

Key concepts

Reduction of Order

The "reduction of order" technique is a powerful method used to find a second solution to a second-order linear differential equation when one solution is already known. Imagine you know one solution of a homogeneous linear differential equation, say \(y_1\). Your goal is to find another solution, \(y_2\), that is linearly independent from \(y_1\). This is where reduction of order comes into play.

The idea is to express \(y_2\) in the form \(y_2 = v(x)y_1\), where \(v(x)\) is a function to be determined. By plugging this form into the original differential equation, you can simplify and solve for \(v(x)\) and thereby find another solution, \(y_2\). This method works because it reduces the complexity by leveraging a known solution and helps you avoid starting from scratch.

Complementary Equation

Finding a "complementary equation" is a fundamental step when solving linear differential equations. A complementary equation is actually the homogeneous version of the differential equation you are attempting to solve. Let's say the original equation is:\[ y'' + p(x)y' + q(x)y = g(x) \]To find the complementary part, you consider:\[ y'' + p(x)y' + q(x)y = 0 \]

Its solutions are crucial because they provide the homogeneous solutions that form the general solution when combined with a particular solution of the non-homogeneous equation. In simpler terms, solving the complementary equation gives you the building blocks to construct the full response of the system described by the differential equation.

Linearly Independent Solutions

"Linearly independent solutions" are a key concept for solving differential equations. They are distinct solutions that cannot be expressed as a constant multiple of one another. In other words, one solution cannot be derived simply by multiplying the other by a number.

Why is this important? Because for a second-order linear differential equation, you need two solutions to form the "general solution." These solutions must be linearly independent to ensure the comprehensive capture of behavior described by the equation. If you have one known solution, using the reduction of order to find another independent solution ensures that the general solution is complete and accurate.

Fundamental Set of Solutions

A "fundamental set of solutions" refers to a set of solutions to a differential equation that can be used to express any solution of that equation. When dealing with second-order linear differential equations, a fundamental set consists of two linearly independent solutions. Given these two solutions, you can express the general solution as a linear combination of them.

Mathematically, if you have solutions \(y_1\) and \(y_2\), then the general solution is:\[ y(x) = C_1y_1 + C_2y_2 \]where \(C_1\) and \(C_2\) are arbitrary constants. These constants can be adjusted based on initial conditions or boundary conditions to find a particular solution that fits specific criteria. This flexibility makes the fundamental set extremely valuable when working with differential equations.