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Home / Find Linearly Independent Solutions to the Homogeneous System

Find Linearly Independent Solutions to the Homogeneous System

Second Order Linear Equations

A second order linear differential equation can be written as

\begin{displaymath} y'' + p(x) y' + q(x) y = g(x) \end{displaymath}


where $p$, $q$ and $g$ are arbitrary functions, and $x$ is the independent variable. Particularly important are the constant-coefficient equations, where $p$ and $q$ (but not necessarily $g$) are constants, and the homogeneous equations, where $g$ is $0$.

The theory of linear differential equations is closely related to the theory of systems of linear equations in linear algebra. As a result of this theory, we will be able to say a lot about the structure of the solutions. Specifically, our main results will be the following:

  1. To find the general solution of a second order homogeneous linear equation, we need to find two linearly independent solutions $y_1$ and $y_2$. All the solutions are linear combinations of these: $y = c_1 y_1 + c_2 y_2$.
  2. To find the general solution of a second order non-homogeneous linear equation, we need to find one solution $y_p$ of it and two linearly independent solutions $y_1$ and $y_2$ of the corresponding homogeneous equation. All the solutions are of the form $y = y_p + c_1 y_1 + c_2 y_2$.

Example: One solution of the non-homogeneous differential equation $y'' + y = x$ is $y = x$. Two linearly independent solutions of the homogeneous equation $y'' + y = 0$ are $y = \cos x$ and $y = \sin x$. We're not concerned right now with how to obtain these solutions, but it's easy to check that they are solutions. The conclusion will be that the general solution of $y'' + y = x$ is $y = x + c_1 \cos x + c_2 \sin x$.


To make the differential equation look more like linear algebra, we define a linear operator:

\begin{displaymath} Ty = y'' + p y' + q y \end{displaymath}


This is a linear transformation in the sense of linear algebra, but the ``vectors'' it operates on are functions. That is, if $f$ is a twice-differentiable function then $Tf$ is the function

\begin{displaymath}Tf(x) = f''(x) + p(x) f'(x) + q(x) f(x) \end{displaymath}


The fact that $T$ is a linear operator means that for any twice-differentiable functions $f_1$ and $f_2$ and constants $c_1$ and $c_2$,

\begin{displaymath}T(c_1 f_1 + c_2 f_2) = c_1 Tf_1 + c_2 Tf_2 \end{displaymath}


Using the linear operator $T$, the second-order linear differential equation is written $Ty = g$. This shares the following properties with the matrix equation $Ax = b$:


Theorem: Suppose $y_p$ is one solution of the equation $Ty = g$. Then the solutions of $Ty = g$ consist of all functions of the form $y_p + y_h$ where $y_h$ is a solution of the homogeneous equation $T y = 0$. The solutions of the homogeneous equation form a vector space. Proof: If $y_h$ is any solution of the homogeneous equation, then $T(y_p + y_h) = T y_p + T y_h = g + 0 = g$. Conversely if $y$ is any solution of $Ty = g$, then $T(y - y_p) = Ty - T y_p = g - g = 0$ so $y = y_p + (y-y_p)$ with $y - y_p$ a solution of the homogeneous equation.

To show the solutions of the homogenous equation $T y = 0$ form a vector space, we need to show that if $y_1$ and $y_2$ are any solutions of $T y = 0$ and $c_1$ and $c_2$ are constants, then $c_1 y_1 + c_2 y_2$ is also a solution. But this is easy since

\begin{displaymath}T(c_1 y_1 + c_2 y_2) = c_1 Ty_1 + c_2 Ty_2 = 0\end{displaymath}


As initial conditions for a second order equation, we need to specify both the value of $y$ and its derivative at some point $y_0$. We then have the following Existence and Uniqueness Theorem for second order linear equations:


Theorem: Suppose $p(x)$, $q(x)$ and $g(x)$ are continuous on some interval $a \le x \le b$. Given any $x_0$ with $a \le x_0 \le b$ and any real numbers $y_0$ and $y'_0$, there is exactly one solution of the differential equation $Ty = g$ with initial conditions $y(x_0) = y_0$, $y'(x_0) = y'_0$, defined for $a \le x \le b$.

(We will not prove this)

Example: Consider again $y'' + y = x$. We know that $y = x + c_1 \cos x + c_2 \sin x$ is a solution for any constants $c_1$ and $c_2$. We can satisfy any initial conditions at $x_0 = 0$ using a function of this form: $y(0) = c_1$ and $y'(0) = 1 + c_2$ so we would take $c_1 = y_0$ and $c_2 = y'_0 - 1$. By the uniqueness part of the Existence and Uniqueness Theorem, this gives us the only solution satisfying those initial conditions. Since every solution must satisfy some initial conditions at $x_0$, we can conclude that all solutions are of the form $y = x + c_1 \cos x + c_2 \sin x$.

In our example, the solutions of the homogeneous equation are all functions of the form $c_1 \cos x + c_2 \sin x$. These form a vector space. It is in fact a two-dimensional vector space: every one of these functions is written in exactly one way as a linear combination of the two functions $\cos x$ and $\sin x$ which form a basis of the space. They are known as a fundamental set of solutions. Recall from linear algebra:


Definition: A set of vectors $\{ v_1, v_2, \ldots v_n \}$ is linearly independent if the only way to write a linear combination $c_1 v_1 + c_2 v_2 + \ldots + c_n v_n = 0$ is with all the scalars $c_i = 0$.

Note that in our case, the vectors are functions and the scalars are constants; ``$f=0$'' means $f(x)=0$ for all $x$. For a set of two functions, linear independence is easy to check: they are linearly independent unless one is a constant multiple of the other.


Definition: A basis of the vector space of solutions of a second order homogeneous linear equation is called a fundamental set of solutions of the equation.


Theorem: The vector space of solutions of a second order homogeneous linear equation is two-dimensional. Thus a fundamental set of solutions of the equation consists of two linearly independent solutions.

Proof: This is a consequence of the Existence and Uniqueness Theorem. Take any $x_0$ in the interval where $p(x)$ and $q(x)$ are defined. Let $y_1$ be the solution of the equation with initial conditions $y(x_0) = 1$, $y'(x_0) = 0$. Let $y_2$ be the solution with initial condition $y(x_0) = 0$, $y'(x_0) = 1$. These are linearly independent (any constant multiple of $y_1$ has $y'(x_0) = 0$, and any constant multiple of $y_2$ has $y(x_0) = 0$). To satisfy any initial conditions at $x_0$, say $y(x_0) = a$, $y'(x_0) = b$, we can take $y = a y_1 + b y_2$. Thus any initial conditions at $x_0$ can be obtained with a linear combination of $y_1$ and $y_2$. Since every solution satisfies some initial conditions at $x_0$, every solution is a linear combination of $y_1$ and $y_2$. So $\{y_1, y_2\}$ is a basis of the space of solutions. Since this basis has two elements, the space is two-dimensional.


The two results we stated at the beginning are now clear. For a homogeneous equation, we need to find a fundamental set of solutions, which consists of two linearly independent solutions. This constitutes a basis of the space of solutions of the homogeneous equation. For a non-homogeneous equation, we need one solution of it and a basis of solutions, again consisting of two linearly independent solutions, of the homogeneous equation.


Now we have the general structure of the solutions. What we have to do next is find those solutions.

  • About this document ...

Robert Israel
2002-02-07

Find Linearly Independent Solutions to the Homogeneous System

Source: https://www.math.ubc.ca/~israel/m215/lin2o/lin2o.html