A second-order differential equation has $y''$ as its highest derivative; it may also contain $y'$ and $y$. For a linear equation, the normalized form is

$$y''+p(t)y'+q(t)y=g(t).$$

Here $p$, $q$, and $g$ are known functions of the independent variable $t$. The equation is homogeneous when $g(t)=0$ throughout the interval under consideration:

$$y''+p(t)y'+q(t)y=0.$$

If $g$ is not identically zero, the equation is nonhomogeneous. The function $g$ is the input or forcing term.

Because the equation is second order, an initial-value problem normally needs two pieces of information at the same time $t_0$:

$$y(t_0)=y_0, \qquad y'(t_0)=y'_0.$$

The first condition fixes the starting value. The second fixes the starting slope.

Homogeneous equations with constant coefficients

Start with

$$ay''+by'+cy=0,$$

where $a$, $b$, and $c$ are constants. We assume $a\ne0$ so that this is genuinely a second-order equation. If $a=0$ but $b\ne0$, it reduces to the first-order equation

$$y'=-\frac cb y,$$

whose solutions are $y(t)=Ce^{-(c/b)t}$.

Why exponential trial functions work

For $a\ne0$, we want a function whose derivatives keep the same shape, so all three terms can share a common factor. Exponentials do exactly that. Try

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

where $r$ is a constant that we still need to find. Exponentials are useful here because differentiating them only introduces powers of $r$:

$$y'(t)=re^{rt}, \qquad y''(t)=r^2e^{rt}.$$

Put these expressions into the differential equation:

$$ar^2e^{rt}+bre^{rt}+ce^{rt}=0.$$

Factor out the common exponential:

$$e^{rt}\bigl(ar^2+br+c\bigr)=0.$$

The exponential $e^{rt}$ is never zero, so the polynomial factor must vanish:

$$\boxed{ar^2+br+c=0}.$$

The polynomial

$$P(r)=ar^2+br+c$$

is the characteristic polynomial, and $P(r)=0$ is the characteristic equation. Each real root gives a real exponential solution. Repeated or non-real roots require an extra step, so the first example uses two distinct real roots.

Example: two distinct real roots

Consider

$$y''+5y'+6y=0.$$

Its characteristic equation is

$$r^2+5r+6=0.$$

Factor it:

$$(r+2)(r+3)=0.$$

The roots are

$$r_1=-2, \qquad r_2=-3,$$

so we get two solutions:

$$y_1(t)=e^{-2t}, \qquad y_2(t)=e^{-3t}.$$

The principle of superposition

For the homogeneous equation

$$y''+p(t)y'+q(t)y=0,$$

any linear combination of solutions is another solution. This is the principle of superposition: if $y_1$ and $y_2$ solve the equation, then

$$y(t)=C_1y_1(t)+C_2y_2(t)$$

also solves it for every $C_1,C_2\in\mathbb R$.

The reason is the linearity of differentiation. Define the differential operator

$$L[y]=y''+p(t)y'+q(t)y.$$

Then

$$\begin{aligned} L[C_1y_1+C_2y_2] &=C_1L[y_1]+C_2L[y_2]\\ &=C_1(0)+C_2(0)\\ &=0. \end{aligned}$$

This argument depends on the right side being zero. If $L[y_1]=g$ and $L[y_2]=g$, then

$$L[C_1y_1+C_2y_2]=(C_1+C_2)g,$$

which is not generally equal to $g$. Arbitrary superposition is therefore a property of the associated homogeneous equation, not of a nonhomogeneous equation with fixed forcing.

A complete initial-value problem

Return to

$$y''+5y'+6y=0.$$

Superposition gives the family

$$y(t)=C_1e^{-2t}+C_2e^{-3t}.$$

Now impose

$$y(0)=2, \qquad y'(0)=3.$$

Evaluating the solution and its derivative at $t=0$ gives

$$C_1+C_2=2,$$

and

$$-2C_1-3C_2=3.$$

The same system can be written as

$$\begin{pmatrix} 1&1\\ -2&-3 \end{pmatrix} \begin{pmatrix} C_1\\ C_2 \end{pmatrix} = \begin{pmatrix} 2\\ 3 \end{pmatrix}.$$

Its determinant is $-1$, so the system has one solution:

$$C_1=9, \qquad C_2=-7.$$

Therefore the unique solution of the IVP is

$$\boxed{y(t)=9e^{-2t}-7e^{-3t}}.$$

A quick check confirms both initial conditions:

$$\begin{aligned} y(0)&=9-7=2,\\ y'(t)&=-18e^{-2t}+21e^{-3t},\\ y'(0)&=-18+21=3. \end{aligned}$$

The initial slope lifts the solution before both modes decay

The positive initial slope explains why the curve rises at first. Both exponential terms decay as $t$ increases, so the solution eventually approaches zero.

Existence and uniqueness on an interval

Consider the linear IVP

$$\begin{aligned} y''+p(t)y'+q(t)y&=g(t),\\ y(t_0)&=y_0,\\ y'(t_0)&=y'_0. \end{aligned}$$

Let $I$ be an open interval with $t_0\in I$. If $p$, $q$, and $g$ are continuous on the interval $I$, then the IVP has exactly one solution on the entire interval $I$:

$$y:I\to\mathbb R.$$

This is stronger than a merely local statement. The solution does not stop at an interior point where the normalized coefficients remain continuous. The interval ends only where the chosen coefficient interval ends; it need not be all of $\mathbb R$.

If an equation is initially written as

$$a_2(t)y''+a_1(t)y'+a_0(t)y=h(t),$$

first divide by $a_2(t)$. The normalized coefficients are

$$p(t)=\frac{a_1(t)}{a_2(t)}, \qquad q(t)=\frac{a_0(t)}{a_2(t)}, \qquad g(t)=\frac{h(t)}{a_2(t)}.$$

The theorem applies on an interval where $a_2(t)\ne0$ and these three quotients are continuous. A zero of $a_2$ can split the real line into separate coefficient intervals.

The Wronskian and fundamental solutions

The Wronskian answers a concrete question: do two solutions provide enough freedom to match any starting value and slope? Given two differentiable functions $f$ and $g$, their Wronskian is

$$\begin{aligned} W[f,g](t) &=\det \begin{pmatrix} f(t)&g(t)\\ f'(t)&g'(t) \end{pmatrix}\\ &=f(t)g'(t)-f'(t)g(t). \end{aligned}$$

The Wronskian tells us whether two solution functions provide independent initial data. Suppose $y_1$ and $y_2$ solve the same homogeneous second-order linear equation on $I$, with $p$ and $q$ continuous there. To make

$$y=C_1y_1+C_2y_2$$

satisfy prescribed values at $t_0$, the constants must solve

$$\begin{pmatrix} y_1(t_0)&y_2(t_0)\\ y_1'(t_0)&y_2'(t_0) \end{pmatrix} \begin{pmatrix} C_1\\ C_2 \end{pmatrix} = \begin{pmatrix} y_0\\ y'_0 \end{pmatrix}.$$

The determinant of the coefficient matrix is exactly

$$W[y_1,y_2](t_0).$$

The matrix can match every pair of initial values if and only if

$$\boxed{W[y_1,y_2](t_0)\ne0}.$$

When this condition holds, the matrix is invertible, so every pair of initial values determines one pair of constants $C_1,C_2$. The existence-and-uniqueness theorem then shows that this linear combination is the only solution with those initial values.

In that case, $\{y_1,y_2\}$ is called a fundamental set of solutions on $I$, and

$$\boxed{y(t)=C_1y_1(t)+C_2y_2(t)}$$

is the general solution of the homogeneous equation: every solution on $I$ appears exactly once for a choice of $C_1$ and $C_2$.

For the example $y_1=e^{-2t}$ and $y_2=e^{-3t}$,

$$\begin{aligned} W[y_1,y_2](t) &=e^{-2t}\bigl(-3e^{-3t}\bigr) -\bigl(-2e^{-2t}\bigr)e^{-3t}\\ &=-e^{-5t}. \end{aligned}$$

This never vanishes, so the two exponentials form a fundamental set of solutions on $\mathbb R$.