Second-Order Linear Equations

Definition A second-order linear differential equation has the form

with $P$, $Q$, $R$ and $G$ functions of $t$ alone. It is homogeneous when $G(t) = 0$ for all $t$. On an interval where $P(t) \ne 0$ we may divide by $P$ and get the normalized form

Points where $P(t) = 0$ are singular points, and the theory below is applied on intervals that avoid them.

Definition Let $p$, $q$ and $g$ be continuous on an open interval $I$, and let $t_0$ be in $I$. For any real numbers $y_0$ and $y_1$, the initial value problem

has exactly one solution, and that solution exists on the whole interval $I$.

Definition If $y_1$ and $y_2$ solve the homogeneous equation $y'' + p(t)\,y' + q(t)\,y = 0$ on an interval $I$, then so does $c_1 y_1 + c_2 y_2$ for all constants $c_1$ and $c_2$. This follows from linearity of differentiation. The principle fails for nonhomogeneous equations, since there the right side would be multiplied by $c_1 + c_2$.

Definition Two functions $y_1$ and $y_2$ are linearly independent on an interval $I$ if the only constants with $c_1 y_1(t) + c_2 y_2(t) = 0$ for every $t$ in $I$ are $c_1 = c_2 = 0$. Equivalently, neither is a constant multiple of the other on $I$. If $p$ and $q$ are continuous on $I$ and $y_1$, $y_2$ are linearly independent solutions of the homogeneous equation on $I$, they form a fundamental set, and every solution on $I$ has the form

Definition The Wronskian of two differentiable functions is

If $y_1$ and $y_2$ solve $y'' + p(t)\,y' + q(t)\,y = 0$ with $p$ and $q$ continuous on $I$, then Abel’s formula holds,

so $W$ is either zero at every point of $I$ or zero at no point of $I$. The two solutions are independent on $I$ exactly when $W \ne 0$. For any two differentiable functions, $W \ne 0$ at a single point already forces independence. The converse needs the functions to be solutions. The functions $t^2$ and $t \lvert t \rvert$ on $\mathbb{R}$ have $W = 0$ everywhere, since the derivative of $t \lvert t \rvert$ is $2\lvert t \rvert$, yet they are independent on $\mathbb{R}$, because $t \lvert t \rvert$ equals $t^2$ for $t > 0$ and $-t^2$ for $t < 0$. So they cannot both solve one equation of this form with $p$ and $q$ continuous on $\mathbb{R}$.

Definition For the equation $a y'' + b y' + c y = 0$ with real constants $a$, $b$, $c$ and $a \ne 0$, the characteristic equation is

Substituting $y = e^{rt}$ gives $(a r^2 + b r + c)\,e^{rt} = 0$, and since $e^{rt} \ne 0$, the exponential $e^{rt}$ is a solution exactly when $r$ is a root. The coefficients are constant, hence continuous on all of $\mathbb{R}$, so every solution is defined for all real $t$.

Definition Let $a \ne 0$, $b$, $c$ be real.

• If $b^2 - 4ac > 0$, the roots $r_1 \ne r_2$ are real and the general solution is $y = c_1 e^{r_1 t} + c_2 e^{r_2 t}$. The Wronskian is $(r_2 - r_1)\,e^{(r_1 + r_2)t} \ne 0$.
• If $b^2 - 4ac = 0$, there is one root $r = -b/(2a)$, and the general solution is $y = (c_1 + c_2 t)\,e^{rt}$. The second solution $t e^{rt}$ works because $2ar + b = 0$, and the Wronskian is $e^{2rt} \ne 0$.

Definition If $b^2 - 4ac < 0$, the roots are $r = \alpha \pm i\beta$ with

Since the coefficients are real, the real and imaginary parts of the complex solution $e^{(\alpha + i\beta)t}$ are real solutions, and the general real solution is

with Wronskian $\beta e^{2\alpha t} \ne 0$.

Definition An initial value problem prescribes $y(t_0)$ and $y'(t_0)$ at one point, and under the continuity hypotheses above it always has exactly one solution. A boundary value problem prescribes values at two different points, such as $y(t_0) = A$ and $y(t_1) = B$ with $t_0 \ne t_1$. A boundary value problem may have exactly one solution, no solution, or infinitely many, and the general solution must be checked against both conditions.