Higher-Order Linear ODEs

Higher-Order Linear ODEs

As we have already noted, second-order equations arise very often in applications.

However, higher-order equations do come up from time to time.

For example, under certain conditions, transversal vibrations in a beam can be described by the homogeneous, linear, constant-coefficient fourth-order equation of the form $$ ay^{(4)}-ky=0 $$ Therefore, we extend the theory of second-order linear equations to higher-order linear equations.

$n$th-Order Initial-Value Problems

An $n$th-order initial value problem where we solve for a function $y$ will have $n$ initial conditions. $$ y(x_0)=a_0, \,\,\,\,y'(x_0)=a_1,\,\,\,\,y''(x_0)=a_2,\,\,\,\,\ldots,\,\,\,\,y^{(n-1)}(x_0)=a_{n-1} $$

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 $n$th-order equations have $n$ constants, $C_1,$ $C_2,$ $\ldots,$ $C_n.$

$n$th-Order Linear ODEs

An $n$th-order linear differential equation has the form $$A_n(x)y^{(n)}+A_{n-1}(x)y^{(n-1)}+\cdots+A_2(x)y''+A_1(x)y'+A_0(x)y=F(x)$$ Dividing through by $A(x),$ we may, without loss of generality, consider solving instead $$y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=f(x)$$ The remainder of this section is devoted to solving the second equation.

Homogeneous $n$th-Order Linear ODEs

A $n$th-order linear differential equation is called homogeneous if $f(x)=0.$

That is, when our equation has the form $$ y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=0 $$ The first step in finding all solutions to the general equation will involve solving the above homogeneous equation first.

Recall: 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."

Recall: 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 Differential Operators: An Example

The operator $L$ defined by $$ Ly=y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y $$ is an example of a linear differential operator.

Fun Fact

Using the linearity of derivatives, it's easy to prove an important theorem about solutions to $n$th-order linear ODEs!

Superposition Theorem

Suppose that $y_1,$ $y_2,$ $\ldots$ $y_n$ are solutions to the $n$th-order linear homogeneous equation $$ y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=0 $$ Then $$ y=C_1y_1+C_2y_2+\cdots+C_ny_n $$ is also a solution.

Since $y_1,$ $y_2,$ $\ldots,$ $y_n$ are solutions, we have $Ly_1=0,$ $Ly_2=0,$ $\ldots,$ $Ly_n=0.$

Then $$ \begin{array}{lll} &\displaystyle y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y\displaystyle&\\ &=Ly &\mbox{}\\ \displaystyle &\displaystyle=L(C_1y_1+C_2y_2+\cdots+C_ny_n) &\mbox{}\\ \displaystyle &\displaystyle=C_1Ly_1+C_2Ly_2+\cdots+C_nLy_n &\mbox{by linearity of $L$}\\ \displaystyle &\displaystyle=C_1\cdot 0+C_2\cdot 0+\cdots+C_n\cdot 0 &\mbox{}\\ \displaystyle &\displaystyle=0 &\mbox{}\\ \end{array} $$

Huge Observation

For an $n$th-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).

Big Question

We've seen that if $y_1$ and $y_2$ are solutions to a $n$th-order linear homogeneous equation, then $ y=C_1y_1+C_2y_2+\cdots+C_ny_n $ is also a solution.

Do all linear combinations of $y_1,$ $y_2,$ $\ldots,$ $y_n$ give us all solutions?

Big Answer

Yes! (Almost)

To get all solutions, $y_1,$ $y_2,$ $\ldots,$ $y_n$ must be linearly independent.

Linear independence guarantees that $y_1,$ $y_2,$ $\ldots,$ $y_n$ are not constant multiples of one another.

Linear Independence

Let $C_1,$ $C_2,$ $\ldots,$ $C_n$ be constants.

$n$ functions $y_1,$ $y_2,$ $\ldots,$ $y_n$ are linearly independent if $$ C_1y_1+C_2y_2+\cdots+C_ny_n=0 $$ implies $$ C_1=0 \,\,\,\,C_2=0,\ldots,\,\,\,\, C_n=0 $$ Again, this is formal way to say that $y_i$ cannot be a constant multiple of $y_j$ if $i \neq j.$

Huge Result

Let $p$ and $q$ be continuous functions. Let $y_1,$ $y_2,$ $\ldots,$ $y_n$ be $n$ linearly independent solutions to the homogeneous equation $$ y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=0 $$ Then every other solution is of the form $$ y=C_1y_1+C_2y_2+\cdots+C_ny_n $$ That is, $y=C_1y_1+C_2y_2+\cdots+C_ny_n$ is the general solution. (There aren't any others!)

Vocab

Let $p$ and $q$ be continuous functions and let $y_1,$ $y_2,$ $\ldots,$ $y_n$ be $n$ linearly independent solutions to the homogeneous equation $$ y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=0 $$ Since any solution can be written in the form $$ y=C_1y_1+C_2y_2+\cdots+C_ny_n $$ the collection $\{y_1,y_2,\ldots,y_n\}$ is called a fundamental solution set of the equation.

Existence and Uniqueness for $n$th-Order Linear Equations

Suppose $p_{n-1},$ $\ldots,$ $p_0,$ and $f$ are continuous functions on some interval $I$ with $x_0\in I,$ and $b_0,$ $b_1,$ $\ldots,$ $b_{n-1}$ are constants. Then the equation $$y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=f(x)$$ has exactly one solution $y(x)$ defined on the same interval $I$ satisfying the initial conditions $$y(x_0)=a_0, \,\,\,\,y'(x_0)=a_1,\,\,\,\,\ldots,\,\,\,\, y^{(n-1)}(x_0)=a_{n-1}$$

Example

Consider the third-order linear equation $$ x^3y'''+x^2y''-2xy'+2y=0 $$ For the initial conditions $y(1)=0,$ $y'(1)=1,$ and $y''(1)=3,$ what is the largest interval over which the solution is guaranteed to exist and be unique?

Getting $x^3y'''+x^2y''-2xy'+2y=0$ into the form of the Existence and Uniqueness Theorem, $$ y'''+\frac{1}{x}y''-\frac{2}{x^2}y'+\frac{2}{x^3}y=0 $$ Since the coefficients on the zero, first, and second order terms are discontinuous at $x_0=0,$ the largest interval over which the solution $y$ satisfying the initial conditions at $x_0=1$ is guaranteed to exist and be unique is $I=(0,\infty).$

The Wronskian

Let $y_1,$ $y_2,$ $\ldots,$ $y_n$ be $n$ solutions to the homogeneous equation $$ y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=0 $$ The Wronskian of $y_1,$ $y_2,$ $\ldots,$ $y_n$ is $$ W[y_1,y_2,\ldots,y_n]= \left| \begin{array}{cccc} y_1 & y_2 &\ldots & y_n\\ y_1'& y_2' & \ldots & y_n'\\ \vdots& \vdots & \vdots & \vdots\\ y_1^{(n-1)}& y_2^{(n-1)} & \ldots & y_n^{(n-1)}\\ \end{array} \right| $$

The Wronskian

Let $p_{n-1},$ $\ldots,$ $p_0,$ be continuous functions. Let $y_1,$ $y_2,$ $\ldots,$ $y_n$ be $n$ solutions to the homogeneous equation $$ y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_2(x)y''+p_1(x)y'+p_0(x)y=0 $$ Then $y_1,$ $y_2,$ $\ldots,$ and $y_n$ are linearly independent on an interval $(a,b)$ if and only if $W[y_1,y_2,\ldots,y_n]\neq 0$ on $(a,b).$

Example

The functions $y_1=x,$ $y_2=x^2,$ and $y_3=x^{-1}$ are solutions to the third-order linear equation $$ x^3y'''+x^2y''-2xy'+2y=0 $$ (a) Show that $y_1,$ $y_2,$ and $y_3$ are linearly independent on $(0,\infty)$ directly by using the definition of linear independence.

(b) Show that $y_1,$ $y_2,$ and $y_3$ are linearly independent on $(0,\infty)$ by using the Wronskian.

(c) State the fundamental solution set of the equation on $(0,\infty).$

(d) Find the general solution of the equation on $(0,\infty).$

(a) Suppose $$ C_1x+C_2x^2+C_3 x^{-1}=0 $$ Multiplying both sides by $x,$ $$ C_1x^2+C_2x^3+C_3=0 $$ This would mean that $C_1x^2+C_2x^3$ is a constant value, namely $-C_3,$ for all $x\in(0,\infty).$

The only way this could happen is if $C_1=C_2=0$ which would also require that $C_3=0.$

It follows that $y_1,$ $y_2,$ and $y_3$ are linearly independent on $(0,\infty).$

(b) $$ \begin{array}{lll} \displaystyle W[y_1,y_2,y_3]&\displaystyle=\left|\begin{array}{ccc} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3''\\ \end{array}\right| &\mbox{}\\ \displaystyle &\displaystyle=\left|\begin{array}{ccc} x & x^2 & x^{-1} \\ 1 & 2x & -x^{-2} \\ 0 & 2 & 2x^{-3}\\ \end{array}\right| &\mbox{}\\ \displaystyle &\displaystyle=x\left|\begin{array}{cc} 2x & -x^{-2} \\ 2 & 2x^{-3} \\ \end{array}\right|-1\cdot \left|\begin{array}{cc} x^2 & x^{-1} \\ 2 & 2x^{-3} \\ \end{array}\right|+0\cdot \left|\begin{array}{cc} x^2 & x^{-1} \\ 2x & -x^{-2} \\ \end{array}\right| &\mbox{}\\ \displaystyle &\displaystyle=x(4x^{-2}-(-2x^{-2}))-(2x^{-1}-2x^{-1}) &\mbox{}\\ \displaystyle &\displaystyle=x(6x^{-2})-0 &\mbox{}\\ \displaystyle &\displaystyle=6x^{-1}&\mbox{}\\ \end{array} $$ Since $6x^{-1}\gt 0$ for all $x\in (0,\infty),$ the Wronskian is never $0$ on $(0,\infty).$

It follows that $y_1=x,$ $y_2=x^2,$ and $y_3=x^{-1}$ are linearly independent on $(0,\infty).$

(c) The fundamental solution set is $\{x,x^2,x^{-1}\}.$

(d) Since $y_1=x,$ $y_2=x^2,$ and $y_3=x^{-1}$ are linearly independent on $(0,\infty),$ we know that $$ y=C_1x+C_2x^2+C_3 x^{-1} $$ is the general solution to the equation on $(0,\infty).$

Solving $n$th-order Linear Homogeneous ODEs with Constant Coefficients

We now extend the theory of homogeneous, second-order equations with constant coefficients to $n$th order. $$ a_n y^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_2y''+a_1 y'+a_0y=0 $$ where $a_n,$ $a_{n-1},$ $\ldots,$ $a_1,$ $a_0$ are constants.

In like manner to second-order equations, we begin by assuming a solution of the form $y=e^{rx}.$

The Characteristic Equation

If $y=e^{rx},$ then $$ \begin{array}{lll} &\displaystyle a_ny^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_2y''+a_1y'+a_0y=0&\mbox{}\\ \implies &\displaystyle a_n(e^{rx})^{(n)}+a_{n-1}(e^{rx})^{(n-1)}+\cdots+a_2(e^{rx})''+a_1(e^{rx})'+a_0e^{rx}=0&\mbox{}\\ \implies &\displaystyle a_nr^n e^{rx}+a_{n-1}r^{n-1}e^{rx}+\cdots+a_2r^2 e^{rx}+a_1re^{rx}+a_0e^{rx}=0&\mbox{}\\ \implies &\displaystyle \left(a_nr^n+a_{n-1}r^{n-1}+\cdots+a_2r^2+a_1r+a_0\right)e^{rx}=0&\mbox{}\\ \implies &\displaystyle a_nr^n+a_{n-1}r^{n-1}+\cdots+a_2r^2+a_1r+a_0=0&\mbox{since $e^{rx}$ can never be $0$}\\ \end{array} $$

Case $1$: $n$ Distinct Real Roots

If the equation $$ a_n y^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_2y''+a_1 y'+a_0y=0 $$ has a characteristic equation with $n$ distinct, real roots $r_1,$ $r_2,$ $\ldots,$ $r_n,$ then the general solution to the equation is $$ y=C_1e^{r_1x}+C_2e^{r_2x}+\cdots+C_ne^{r_nx} $$

Example

Find the general solution to the equation $$ y'''-2y''-5y'+6y=0 $$

The characteristic equation is $$ r^3-2r^2-5r+6=0 $$ By the Rational Root Theorem, the possible rational solutions are $\pm 1$ $\pm 2,$ $\pm 3,$ and $\pm 6.$

Trying these, we see that $r=1$ solves the equation. Thus, $$ r^3-2r^2-5r+6=(r-1)Q(x) $$ where $Q(x)$ is a quadratic polynomial.

Performing the polynomial long division, $$ r-1)\overline{r^3-2r^2-5r+6} $$

we get a quotient of $r^2-r-6$ and a remainder of $0.$

Therefore, $$ r^3-2r^2-5r+6=(r-1)(r^2-r-6)=(r-1)(r+2)(r-3)=0. $$ Our $3$ distinct real roots are $r_1=1,$ $r_2=-2,$ and $r_3=3$ so that our general solution is $$ y=C_1e^{x}+C_2e^{-2x}+C_3e^{3x} $$

Complex Roots

It's a fact that complex roots of polynomials always occur in conjugate pairs $\alpha \pm \beta i.$

That is, if $\alpha + \beta i$ is a root, then it's guaranteed that its conjugate $\alpha - \beta i$ is also a root, and vice versa.

So, if $\alpha \pm \beta i$ are complex roots to the characteristic equation, then $$ e^{\alpha x}\cos(\beta x) \,\,\,\,\mbox{and}\,\,\,\, e^{\alpha x}\sin(\beta x) $$ are two linearly independent solutions to the equation.

This allows us to state Case $2.$

Case $2$: $n$ Distinct Roots.

If the equation $$ a_n y^{(n)}+a_{n-1}y^{(n-1)}+\cdots+a_2y''+a_1 y'+a_0y=0 $$ has a characteristic equation with $n$ distinct roots $r_1,$ $r_2,$ $\ldots,$ $r_n,$ with the possiblility of complex conjugate pairs, then the general solution to the equation is $$ y=C_1y_1+C_2y_2+\cdots+C_ny_n $$ where $$ y_j= e^{r_j x}\,\,\,\,\mbox{if}\,\,\,\,r_j \in \mathbb{R} $$ and $$ y_j=e^{\alpha x}\cos(\beta x) \,\,\,\,\mbox{and}\,\,\,\, y_{j+1}=e^{\alpha x}\sin(\beta x) \,\,\,\,\mbox{if}\,\,\,\, r_j=\alpha + \beta i \,\,\,\,\mbox{and}\,\,\,\,r_{j+1}=\alpha - \beta i $$

Example

Find the general solution to the equation $$ y'''+y''+3y'-5y=0 $$

The characteristic equation is $$ r^3+r^2+3r-5=0 $$ By the Rational Root Theorem, the possible rational solutions are $\pm 1$ and $\pm 5.$

Trying these, we see that $r=1$ solves the equation. Thus, $$ r^3+r^2+3r-5=(r-1)Q(x) $$ where $Q(x)$ is a quadratic polynomial.

Performing the polynomial long division, $$ r-1)\overline{r^3+r^2+3r-5} $$

we get a quotient of $r^2+2r+5$ and a remainder of $0.$

Therefore, $$ r^3+r^2+3r-5=(r-1)(r^2+2r+5)=0. $$ Solving $r^2+2r+5=0$ by the quadratic formula gives $r_2=-1+2i$ and $r_3=-1-2i.$

Our $3$ distinct roots are $r_1=1,$ $r_2=-1+2i,$ and $r_3=-1-2i$ so that our general solution is $$ y=C_1e^{x}+C_2e^{-x}\cos(2x)+C_3e^{-x}\sin(2x) $$

Case $3$: Repeated Real Roots

If the characteristic equation has a real root $r$ of multiplicity $m\leq n,$ then $$ e^{rx}, xe^{rx},\ldots,x^{m-1}e^{rx} $$ are $m$ linearly independent solutions.

Example

Find the general solution to the equation $$ y^{(4)}-y'''-3y''+5y'-2y=0 $$

The characteristic equation is $$ r^4-r^3-3r^2+5r-2=0 $$ By the Rational Root Theorem, the possible rational solutions are $\pm 1$ and $\pm 2.$

Trying these, we see that $r=1$ solves the equation. Thus, $$ r^4-r^3-3r^2+5r-2=(r-1)C(x) $$ where $C(x)$ is a cubic polynomial.

Performing the polynomial long division, $$ r-1)\overline{r^4-r^3-3r^2+5r-2} $$

we get a quotient of $r^3-3r+2$ and a remainder of $0.$

Therefore, $$ r^4-r^3-3r^2+5r-2=(r-1)(r^3-3r+2)=0. $$ We now focus on the equation $r^3-3r+2=0.$ Again, by the Rational Root Theorem, the possible rational solutions are $\pm 1$ and $\pm 2.$

Trying these, we see again that $r=1$ solves the equation. Thus, $$ r^4-r^3-3r^2+5r-2=(r-1)(r^3-3r+2)=(r-1)(r-1)Q(x)=(r-1)^2Q(x) $$ where $Q(x)$ is a quadratic polynomial.

Performing the polynomial long division, $$ r-1)\overline{r^2-3r+2} $$

we get a quotient of $r^2+r-2$ and a remainder of $0.$

Thus, $$ \begin{array}{ll} r^4-r^3-3r^2+5r-2&=(r-1)(r^3-3r+2)\\ &=(r-1)(r-1)(r^2+r-2)\\ &=(r-1)(r-1)(r-1)(r+2)\\ &=(r-1)^3(r+2)\\ &=0\\ \end{array} $$ That is, $r=r_1=r_2=r_3=1$ is a repeated real root of multiplicity $3$ and $r_4=-2.$

Our general solution is therefore $$ y=C_1e^{x}+C_2xe^{x}+C_3x^2e^{x}+C_4e^{-2x} $$

Case $4$: Repeated Complex Roots

If $r_j=\alpha+\beta i$ and $r_{j+1}=\alpha-\beta i$ are roots of multiplicity $m,$ then $$ e^{\alpha x}\cos(\beta x),xe^{\alpha x}\cos(\beta x), \ldots, x^{m-1}e^{\alpha x}\cos(\beta x) $$ are $m$ linearly independent solutions, and $$ e^{\alpha x}\sin(\beta x),xe^{\alpha x}\sin(\beta x), \ldots, x^{m-1}e^{\alpha x}\sin(\beta x) $$ are $m$ linearly independent solutions for a total of $2m \leq n$ linearly independent solutions.

Example

Find the general solution to the equation $$ y^{(4)}-4y'''+8y''-8y'+4y=0 $$

The characteristic equation is $$ r^4-4r^3+8r^2-8r+4=0 $$ The Rational Root Theorem, tells us that the possible rational roots are $\pm 1,$ $\pm 2,$ and $\pm 4.$

Of these, none are roots. So, our characteristic equation has no rational roots.

At this point, we would need to resort to trial and error, graphing techniques, numerical methods, or just plain divine inspiration.

It turns our that $$ r^4-4r^3+8r^2-8r+4=(r^2-2r+2)^2=0 $$ Then, finding solutions to $r^2-2r+2=0,$ we have $r=1\pm i.$

Thus, $r_1=1+i$ and $r_2=1-i$ are repeated complex conjugate roots of multiplicity $2.$

Therefore, our general solution is $$ y=C_1e^x\cos x+C_2xe^x\cos x+C_3e^x\sin x+C_4xe^x\sin x $$