A higher-order linear equation follows the same basic pattern as a second-order one: first solve the associated homogeneous equation, then add one particular solution if there is a forcing term. The difference is that an equation of order $n$ needs $n$ independent homogeneous solutions and $n$ initial conditions.

General nth-order linear equations

On an interval $I$, consider the normalized equation

$$y^{(n)}+p_1(t)y^{(n-1)}+\cdots+p_{n-1}(t)y'+p_n(t)y=g(t).$$

Here $y^{(k)}$ means the $k$th derivative of $y$ with respect to $t$. The coefficient of $y^{(n)}$ is $1$, which is why the equation is called normalized. If an equation begins with a nonzero leading coefficient, divide through by it on an interval where it does not vanish before using this form.

The equation is homogeneous when the right side is zero:

$$y^{(n)}+p_1(t)y^{(n-1)}+\cdots+p_{n-1}(t)y'+p_n(t)y=0.$$

When $g$ is not identically zero, the equation is nonhomogeneous, and $g(t)$ is the forcing term.

Initial conditions and the interval theorem

An $n$th-order initial-value problem specifies the value of the solution and its first $n-1$ derivatives at one point $t_0$:

$$y(t_0)=b_0, \quad y'(t_0)=b_1, \quad\ldots,\quad y^{(n-1)}(t_0)=b_{n-1}.$$

Suppose $I$ is an open interval, $t_0\in I$, and

$$p_1,\ldots,p_n,g$$

are continuous on $I$. Then every choice of the $n$ initial values $b_0,\ldots,b_{n-1}$ determines exactly one solution on all of $I$. As in the second-order theorem, the interval matters: a zero of the original leading coefficient or a discontinuity in a normalized coefficient may separate the real line into different valid intervals.

Fundamental sets and the Wronskian

Suppose $y_1,\ldots,y_n$ all solve the associated homogeneous equation on $I$. Their Wronskian is

$$W[y_1,\ldots,y_n](t) = \det \begin{pmatrix} y_1(t)&y_2(t)&\cdots&y_n(t)\\ y_1'(t)&y_2'(t)&\cdots&y_n'(t)\\ \vdots&\vdots&\ddots&\vdots\\ y_1^{(n-1)}(t)&y_2^{(n-1)}(t)&\cdots&y_n^{(n-1)}(t) \end{pmatrix}.$$

Each column records one solution and its first $n-1$ derivatives. If

$$W[y_1,\ldots,y_n](t_0)\ne0$$

at one point $t_0\in I$, the columns are independent. The initial-value theorem then shows that every set of $n$ initial values can be matched by exactly one linear combination

$$c_1y_1+\cdots+c_ny_n.$$

In that case, $\{y_1,\ldots,y_n\}$ is a fundamental set of solutions on $I$, and the homogeneous general solution is

$$\boxed{y_c(t)=c_1y_1(t)+\cdots+c_ny_n(t)}.$$

The subscript $c$ stands for complementary solution. For solutions of the same regular homogeneous linear equation, a Wronskian that is nonzero at one point stays nonzero throughout the interval.

Adding a particular solution

Let $y_p$ be any one particular solution of the nonhomogeneous equation. Then

$$\boxed{ y(t)=y_c(t)+y_p(t) =c_1y_1(t)+\cdots+c_ny_n(t)+y_p(t) }$$

is its general solution.

The reason is linearity. If $L[y]=g$ denotes the nonhomogeneous equation and both $y$ and $y_p$ satisfy it, then

$$L[y-y_p]=L[y]-L[y_p]=g-g=0.$$

So $y-y_p$ must be a homogeneous solution, and the fundamental set expresses it as $c_1y_1+\cdots+c_ny_n$. The $n$ constants in $y_c$ provide exactly the freedom needed to match the $n$ initial conditions. The particular solution adds no new arbitrary constant.

Homogeneous equations with constant coefficients

Now consider

$$a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_{n-1}y'+a_ny=0, \qquad a_0\ne0,$$

where every $a_j$ is constant. Exponential functions are useful because each differentiation only multiplies them by another factor of the same number. Try

$$y(t)=e^{rt}.$$

Then $y^{(k)}=r^ke^{rt}$, so substitution gives

$$e^{rt}\left(a_0r^n+a_1r^{n-1}+\cdots+a_{n-1}r+a_n\right)=0.$$

Because $e^{rt}$ never vanishes, $r$ must satisfy the characteristic equation

$$\boxed{P(r)=a_0r^n+a_1r^{n-1}+\cdots+a_{n-1}r+a_n=0}.$$

The Fundamental Theorem of Algebra says that a degree-$n$ characteristic polynomial has $n$ complex roots when multiplicity is counted. For now, suppose all $n$ roots $r_1,\ldots,r_n$ are real and distinct. Each root gives an exponential solution $e^{r_jt}$.

Why distinct exponentials are independent

Put the roots in the order used for the columns of the Wronskian. For

$$y_j(t)=e^{r_jt},$$

the Wronskian is

$$\begin{aligned} W(t) &= \det \begin{pmatrix} e^{r_1t}&e^{r_2t}&\cdots&e^{r_nt}\\ r_1e^{r_1t}&r_2e^{r_2t}&\cdots&r_ne^{r_nt}\\ \vdots&\vdots&\ddots&\vdots\\ r_1^{n-1}e^{r_1t}&r_2^{n-1}e^{r_2t}&\cdots&r_n^{n-1}e^{r_nt} \end{pmatrix}\\ &=e^{(r_1+\cdots+r_n)t} \det \begin{pmatrix} 1&1&\cdots&1\\ r_1&r_2&\cdots&r_n\\ \vdots&\vdots&\ddots&\vdots\\ r_1^{n-1}&r_2^{n-1}&\cdots&r_n^{n-1} \end{pmatrix}\\ &=e^{(r_1+\cdots+r_n)t} \prod_{1\le i<j\le n}(r_j-r_i). \end{aligned}$$

The last determinant is a Vandermonde determinant. The exponential factor is never zero, and every difference $r_j-r_i$ is nonzero because the roots are distinct. Therefore the exponentials form a fundamental set, and

$$\boxed{y(t)=c_1e^{r_1t}+\cdots+c_ne^{r_nt}}.$$

A fourth-order example

Solve the initial-value problem

$$y^{(4)}+y'''-7y''-y'+6y=0,$$ $$y(0)=1, \qquad y'(0)=0, \qquad y''(0)=-2, \qquad y'''(0)=-1.$$

The characteristic polynomial is

$$P(r)=r^4+r^3-7r^2-r+6.$$

Since $P(1)=0$, divide by $r-1$:

$$P(r)=(r-1)(r^3+2r^2-5r-6).$$

The cubic vanishes at $r=-1$, and the remaining quadratic factors as $r^2+r-6=(r-2)(r+3)$. Hence

$$\boxed{P(r)=(r+3)(r+1)(r-1)(r-2)}.$$

Using the increasing order

$$r_1=-3, \quad r_2=-1, \quad r_3=1, \quad r_4=2,$$

gives the fundamental solutions

$$e^{-3t},\qquad e^{-t},\qquad e^t,\qquad e^{2t}.$$

At $t=0$, their Wronskian matrix is

$$\begin{pmatrix} 1&1&1&1\\ -3&-1&1&2\\ 9&1&1&4\\ -27&-1&1&8 \end{pmatrix}.$$

Its determinant is the Vandermonde product

$$\prod_{i<j}(r_j-r_i)=240\ne0,$$

so the four exponentials are independent. Write

$$y(t)=c_1e^{-3t}+c_2e^{-t}+c_3e^t+c_4e^{2t}.$$

Substituting the four initial conditions gives

$$\begin{pmatrix} 1&1&1&1\\ -3&-1&1&2\\ 9&1&1&4\\ -27&-1&1&8 \end{pmatrix} \begin{pmatrix} c_1\\c_2\\c_3\\c_4 \end{pmatrix} = \begin{pmatrix} 1\\0\\-2\\-1 \end{pmatrix}.$$

Solving this system gives

$$c_1=-\frac18, \qquad c_2=\frac5{12}, \qquad c_3=\frac{11}{8}, \qquad c_4=-\frac23.$$

Therefore

$$\boxed{ y(t)=-\frac18e^{-3t}+\frac5{12}e^{-t} +\frac{11}{8}e^t-\frac23e^{2t} }.$$

Each exponential satisfies the differential equation because its exponent is a root of $P$. Evaluating the solution and its first three derivatives at $t=0$ gives $1$, $0$, $-2$, and $-1$, respectively, so all four initial conditions are satisfied.