Today we begin our study second-order ODEs
Second-order equations arise very often in applications.
In fact, the author of our textbook notes, "higher order equations do appear from time to time, but generally the world around us is 'second order.'"
Today, we begin to develop the theory of second-order linear equations.
Second-Order Initial-Value Problems
Recall from calculus that we solved the second-order differential equation for a free-falling body without air resistance. $$ \frac{d^2s}{dt^2}=-g $$ Recall that to determine position we needed both an initial position $s(0)=s_0$ AND an initial velocity $s'(0)=v_0.$
This will be the case for any second-order initial value problem.
Second-Order Initial-Value Problems
Any second-order initial value problem where we solve for a function $y$ will have two initial conditions. $$ y(0)=a \,\,\,\,\mbox{and}\,\,\,\, y'(0)=b $$ Our focus today, however, will be developing theory to find general solutions.
More specifically, general solutions to linear second-order ODEs.
Families of Solutions
From our study of first-order equations, we saw that a general solution involves a single constant $C.$
We will soon see that general solutions to second-order equations have two constants, $C_1$ and $C_2.$
Second-Order Linear ODEs
A second-order linear differential equation has the form $$A(x)y''+B(x)y'+C(x)y=F(x)$$ Dividing through by $A(x),$ we may, without loss of generality, consider solving instead $$ y''+p(x)y'+q(x)y=f(x) $$ The remainder of this section is devoted to solving the second equation.
Homogeneous Second-Order Linear ODEs
A second-order linear differential equation is called homogeneous if $f(x)=0.$
That is, when our equation has the form $$ y''+p(x)y'+q(x)y=0 $$ The first step in finding all solutions to the general equation will involve solving the above homogeneous equation first.
This will be our first order of business.
Operators
An operator is a special kind of relation which maps functions to functions.
Informally we may say that "operators eat functions and spit out functions."
Operators: An Example
An example of an operator we all know is the differential operator $D=\displaystyle \frac{d}{dx}.$
$D=\displaystyle \frac{d}{dx}$ "eats" a function $f(x)$ and "spits out" its derivative $f'(x).$
For example, if $y=x^3,$ then $D y=3x^2.$
Linear Operators
Let $y_1$ and $y_2$ be functions. An operator $L$ is called linear if $$ L(y_1+y_2)=Ly_1+Ly_2 $$ and $$ L(cy)=cLy $$ for any constant $c.$
As a consequence of the above, $$ L(C_1 y_1+C_2 y_2)=C_1Ly_1+C_2 Ly_2 $$ where $C_1$ and $C_2$ are constants.
We note that the expression $C_1 y_1+C_2 y_2$ is called a linear combination of $y_1$ and $y_2.$
Linear Operators: An Example
The operator $D=\displaystyle \frac{d}{dx}$ is a linear operator.
Let $y_1=f(x)$ and $y_2=g(x)$ be functions, and $C_1$ and $C_2$ constants.
Then $$ \begin{array}{lll} \displaystyle D(C_1 y_1+C_2 y_2)&\displaystyle=\frac{d}{dx}(C_1f(x)+C_2g(x)) &\mbox{}\\ \displaystyle &\displaystyle=C_1\frac{d}{dx}f(x)+C_2\frac{d}{dx}g(x) &\mbox{}\\ \displaystyle &\displaystyle=C_1Dy_1+C_2Dy_2 &\mbox{}\\ \end{array} $$
Linear Differential Operators: An Example
The operator $L$ defined by $$ Ly=y''+p(x)y'+q(x)y $$ is an example of a linear differential operator.
It's linear because $$ \begin{array}{lll} \displaystyle L(C_1 y_1+C_2 y_2)&\displaystyle=(C_1 y_1+C_2 y_2)''+p(x)(C_1 y_1+C_2 y_2)'+q(x)(C_1 y_1+C_2 y_2) &\mbox{}\\ \displaystyle &\displaystyle=C_1 y_1''+C_2 y_2''+p(x)(C_1 y_1'+C_2 y_2')+q(x)(C_1 y_1+C_2 y_2) &\mbox{by linearity of derivatives}\\ \displaystyle &\displaystyle=C_1 y_1''+C_2 y_2''+C_1 p(x) y_1'+C_2 p(x)y_2'+C_1 q(x)y_1+C_2 q(x) y_2 &\mbox{}\\ \displaystyle &\displaystyle=C_1 y_1''+C_1 p(x) y_1'+C_1 q(x)y_1+C_2 y_2''+C_2 p(x)y_2'+C_2 q(x) y_2 &\mbox{rearrange terms}\\ \displaystyle &\displaystyle=C_1 (y_1''+p(x) y_1'+q(x)y_1)+C_2 (y_2''+p(x)y_2'+q(x) y_2) &\mbox{factor out constants}\\ \displaystyle &\displaystyle=C_1 Ly_1+C_2Ly_2 &\mbox{}\\ \end{array} $$
Fun Fact
Using the linearity of the derivative, we've just essentially proven an important theorem about solutions to second-order linear ODEs!
Superposition Theorem
Suppose that $y_1$ and $y_2$ are solutions to the second-order linear homogeneous equation $$ y''+p(x)y'+q(x)y=0 $$ Then $$ y=C_1y_1+C_2y_2 $$ is also a solution.
Since $y_1$ and $y_2$ are solutions, $Ly_1=0$ and $Ly_2=0.$
Thus, $$ \begin{array}{lll} \displaystyle y''+p(x)y'+q(x)y&\displaystyle=Ly &\mbox{}\\ \displaystyle &\displaystyle=L(C_1y_1+C_2y_2) &\mbox{}\\ \displaystyle &\displaystyle=C_1Ly_1+C_2Ly_2 &\mbox{by linearity of $L$}\\ \displaystyle &\displaystyle=C_1\cdot 0+C_2\cdot 0 &\mbox{}\\ \displaystyle &\displaystyle=0 &\mbox{}\\ \end{array} $$
Thus, $$ \begin{array}{lll} \displaystyle y''+p(x)y'+q(x)y&\displaystyle=Ly &\mbox{}\\ \displaystyle &\displaystyle=L(C_1y_1+C_2y_2) &\mbox{}\\ \displaystyle &\displaystyle=C_1Ly_1+C_2Ly_2 &\mbox{by linearity of $L$}\\ \displaystyle &\displaystyle=C_1\cdot 0+C_2\cdot 0 &\mbox{}\\ \displaystyle &\displaystyle=0 &\mbox{}\\ \end{array} $$
Huge Observation
For a second-order linear homogeneous equation, a linear combination of any two solutions is also a solution.
That is, the collection of solutions to an equation form a "space" of functions that is closed under the operations of addition and multiplication by constants (scalar multiplication).
Example
Consider the linear second-order equation $$ y''-15y'+50y=0 $$ This equation is homogeneous.
It is not difficult to verify that both $y_1=e^{5x}$ and $y_2=e^{10x}$ are solutions.
The superposition theorem says that any linear combination of these solutions, that is $$ y=C_1e^{5x}+C_2e^{10x} $$ is also a solution.
Example
Find a solution to the homogeneous linear second-order equation $$ y''-15y'+50y=0 $$ which satisfies the initial conditions $$ y(0)=0\,\,\,\,\mbox{and}\,\,\,\,y'(0)=5 $$
Recall that the general solution to the equation is
$$
y=C_1e^{5x}+C_2e^{10x}
$$
The first condition, $y(0)=0,$ then gives
$$
\begin{array}{lll}
&\displaystyle C_1e^{5\cdot 0}+C_2e^{10\cdot 0}=0&\mbox{}\\
\implies &\displaystyle C_1+C_2=0&\mbox{}\\
\end{array}
$$
The second condition, $y'(0)=5,$ then gives
$$
\begin{array}{lll}
&\displaystyle C_15e^{5\cdot 0}+C_2 10e^{10\cdot 0}=5&\mbox{}\\
\implies &\displaystyle 5C_1+10C_2=5&\mbox{}\\
\end{array}
$$
To find $C_1$ and $C_2,$ we need to solve the system
$$
\begin{cases}
C_1+C_2=0\\
5C_1+10C_2=5\\
\end{cases}
$$
After some computation, we see that $C_1=-1$ and $C_2=1.$
Thus, a solution to the initial value problem is $$ y=-e^{5x}+e^{10x} $$
Thus, a solution to the initial value problem is $$ y=-e^{5x}+e^{10x} $$
.
Big Question
We've seen that if $y_1$ and $y_2$ are solutions to a second-order linear homogeneous equation, then $ y=C_1y_1+C_2y_2 $ is also a solution.
Do all linear combinations of $y_1$ and $y_2$ give us all solutions?
Big Answer
Yes! (Almost)
To get all solutions, $y_1$ and $y_2$ must be linearly independent.
Linear independence guarantees that $y_1$ and $y_2$ are not constant multiples of one another.
Linear Independence
Let $C_1$ and $C_2$ be constants.
Two functions of $y_1$ and $y_2$ are linearly independent if $$ C_1y_1+C_2y_2=0 $$ implies $$ C_1=0 \,\,\,\,\mbox{and}\,\,\,\, C_2=0 $$ Again, this is formal way to say that $y_1$ cannot be a constant multiple of $y_2$ and vice versa.
Example
Show that $y_1=e^{5x}$ and $y_2=e^{10x}$ are linearly independent.
Suppose we have an arbitrary combination of $y_1$ and $y_2$ which is $0.$
Then
$$
C_1y_1+C_2y_2=0
$$
That is,
$$
C_1e^{5x}+C_2e^{10x}=0
$$
Since $e^{5x}\gt 0$ for all $x,$ we may divide the above equation through by $e^{5x}$ to get
$$
C_1+C_2e^{5x}=0
$$
or,
$$
C_2e^{5x}=-C_1
$$
That is, the above says that $C_2e^{5x}$ must be constant for all $x$.
The only way this could happen is if $C_2=0$ (for any other value, $C_2e^{5x}$ couldn't be a constant function).
But if $C_2=0,$ then by the last equation above, it could only be that $C_1=0.$
It follows that $y_1=e^{5x}$ and $y_2=e^{10x}$ are linearly independent.
The only way this could happen is if $C_2=0$ (for any other value, $C_2e^{5x}$ couldn't be a constant function).
But if $C_2=0,$ then by the last equation above, it could only be that $C_1=0.$
It follows that $y_1=e^{5x}$ and $y_2=e^{10x}$ are linearly independent.
Huge Result
Let $p$ and $q$ be continuous functions. Let $y_1$ and $y_2$ be two linearly independent solutions to the homogeneous equation $$ y''+p(x)y'+q(x)y=0 $$ Then every other solution is of the form $$ y=C_1y_1+C_2y_2 $$ That is, $y=C_1y_1+C_2y_2$ is the general solution. (There aren't any others!)
Vocab
Let $p$ and $q$ be continuous functions and let $y_1$ and $y_2$ be two linearly independent solutions to the homogeneous equation $$ y''+p(x)y'+q(x)y=0 $$ Since any solution can be written in the form $$ y=C_1y_1+C_2y_2 $$ the collection $\{y_1,y_2\}$ is called a fundamental solution set of the equation.
Example
Since $y_1=e^{5x}$ and $y_2=e^{10x}$ are linearly independent solutions to the equation $$ y''-15y'+50y=0 $$ we know that $$ y=C_1e^{5x}+C_2e^{10x} $$ is the general solution to the equation.
The fundamental solution set is $\{e^{5x},e^{10x}\}$
Existence and Uniqueness
Again, Jean Luc asks you recall the vast importance of the conditions for existence and uniqueness.
Existence and Uniqueness for Second-Order Linear Equations
Suppose $p,$ $q,$ and $f$ are continuous functions on some interval $I$ with $a\in I,$ and $b_0,$ $b_1$ are constants. Then the equation $$y''+p(x)y'+q(x)y=f(x)$$ has exactly one solution $y(x)$ defined on the same interval $I$ satisfying the initial conditions $$y(a)=b_0 \,\,\,\,\mbox{and}\,\,\,\, y'(a)=b_1$$
Example
Earlier we saw that $$ y=-e^{5x}+e^{10x} $$ is a solution to the homogeneous linear second-order equation $$ y''-15y'+50y=0 $$ which satisfies the initial conditions $$ y(0)=0\,\,\,\,\mbox{and}\,\,\,\,y'(0)=5 $$ Since the conditions of the Existence and Uniqueness Theorem are satisfied, this is the only solution.
Example
Consider the second-order linear equation $$ (x^2-25)y''+3xy'-2y=\sin x $$ What is the largest interval on which a unique solution exists to the initial value problem with $$ y(1)=1\,\,\,\,\mbox{and}\,\,\,\,y'(1)=1. $$
We first write the equation in standard form.
$$
y''+\frac{3x}{x^2-25}y'-\frac{2}{x^2-25}y=\frac{\sin x}{x^2-25}
$$
We see that $\displaystyle p(x)=\frac{-3x}{x^2-25},$ $\displaystyle q(x)=\frac{-2}{x^2-25},$
and $\displaystyle p(x)=\frac{\sin x}{x^2-25}.$
Since these functions have discontinuities at $x=-5$ and $x=5,$ the largest interval which contains our initial value $x=1$ is $(-5,5).$
Thus $(-5,5)$ is the largest interval on which a unique solution exists to the initial value problem with $$ y(1)=1\,\,\,\,\mbox{and}\,\,\,\,y'(1)=1. $$
Since these functions have discontinuities at $x=-5$ and $x=5,$ the largest interval which contains our initial value $x=1$ is $(-5,5).$
Thus $(-5,5)$ is the largest interval on which a unique solution exists to the initial value problem with $$ y(1)=1\,\,\,\,\mbox{and}\,\,\,\,y'(1)=1. $$
Bonus Content!
The Wronskian
Let $y_1$ and $y_2$ be two solutions to the homogeneous equation $$ y''+p(x)y'+q(x)y=0 $$ The Wronskian of $y_1$ and $y_2$ is $$ W[y_1,y_2]= \left| \begin{array}{cc} y_1 & y_2 \\ y_1'& y_2' \\ \end{array} \right|=y_1y_2'-y_2y_1' $$
The Wronskian
Let $p$ and $q$ be continuous functions. Let $y_1$ and $y_2$ be two solutions to the homogeneous equation $$ y''+p(x)y'+q(x)y=0 $$ Then $y_1$ and $y_2$ are linearly independent on an interval $(a,b)$ if and only if $W[y_1,y_2]\neq 0$ on $(a,b).$
Example
We showed that $y_1=e^{5x}$ and $y_2=e^{10x}$ are linearly independent solutions to the equation $$ y''-15y'+50y=0 $$ We could have used the Wronskian instead. $$ \begin{array}{lll} \displaystyle W[y_1,y_2]&\displaystyle= \left| \begin{array}{cc} y_1 & y_2 \\ y_1'& y_2' \\ \end{array} \right|&\mbox{}\\ \displaystyle &\displaystyle=\left| \begin{array}{cc} e^{5x} & e^{10x} \\ 5e^{5x}& 10e^{10x} \\ \end{array} \right| &\mbox{}\\ \displaystyle &\displaystyle=e^{5x}\cdot 10e^{10x}-e^{10x}\cdot 5e^{5x} &\mbox{}\\ \displaystyle &\displaystyle=5e^{15x} &\mbox{}\\ \displaystyle &\displaystyle\gt 0 &\mbox{}\\ \end{array} $$ Since the Wronskian is nonzero on $(-\infty,\infty),$ we conclude that $y_1=e^{5x}$ and $y_2=e^{10x}$ are linearly independent on $(-\infty,\infty).$