The IVP
$$L(y) = y'' + p(t)y' + q(t)y = g(t), \qquad y(t_0) = y_0,\; y'(t_0) = y_0'$$where $p(t)$, $q(t)$ and $g(t)$ are continuous on an open interval $I$ containing the point $t_0$ always admits a unique solution $y(t) = \phi(t)$ on the interval $I$.
For students who are mathematically matured, we refer to Chapter 6, §8 of E. A. Coddington, "An Introduction to Ordinary Differential Equations" (Englewood Cliffs, NJ: Prentice-Hall, 1961; New York: Dover, 1989) for a proof.
Example. Find the longest interval $I$ in which the IVP
$$(t^3 - 3t)y'' + ty' - (t + 3)y = 0, \quad y(1) = 2,\; y'(1) = 1$$has a unique solution.
Let us rewrite the DE as
$$y'' + \frac{t}{(t^3 - 3t)}y' - \frac{(t+3)}{(t^3 - 3t)}y = 0.$$That is, we have
$$p(t) = \frac{1}{t - 3}, \quad q(t) = \frac{(t+3)}{(t^3 - 3t)}$$where $p$ is continuous over $t < 3$ and $t > 3$, whilst $q$ is continuous over $t < 0$, $0 < t < 3$ and $t > 3$. Our IVP involves the point $t_0 = 1$ that lies in the intersection of $t < 3$ and $0 < t < 3$, i.e., over $I = (0, 3)$ in which the above theorem guarantees the IVP has a unique solution (without solving the DE explicitly).
Example. Consider the IVP
$$y'' + p(t)y' + q(t)y = g(t), \quad y(t_0) = 0,\; y'(t_0) = 0,$$where $p(t)$, $q(t)$ are continuous on an open interval $I$ that contains the point $t_0$.
We observe that $y = \phi(t) \equiv 0$ (for all $t$) satisfies the DE and also $\phi(t_0) = 0 = \phi'(t_0)$. So the last Theorem asserts that at $y = \phi(t) \equiv 0$ is the unique solution for this IVP.
Find the longest intervals in which each of the following IVP is guaranteed to have a solution. There is no need to solve these IVP.
Let $f(t)$ and $g(t)$ be two functions defined on a common domain $I$. We define a new function
$$W(f, g)(t) = f(t)g'(t) - f'(t)g(t)$$called the Wronskian (Józef Maria Hoene-Wroński (1776-1853) Polish Messianist philosopher, mathematician, physicist, inventor, lawyer, and economist.) of $f(t)$ and $g(t)$.
We can rewrite
So the Wronskian is also called the Wronskian determinant of $f(t)$ and $g(t)$.
Example. Recall that $y_1(t) = e^{-2t}$, $y_2(t) = e^{-3t}$ are two solutions of the DE
$$y'' + 5y' + 6 = 0.$$Then
\begin{align} W(f, g)(t) &= \begin{vmatrix} e^{-2t} & e^{-3t} \\ -2e^{-2t} & -3e^{-3t} \end{vmatrix} \\[6pt] &= e^{-2t}(-3e^{-3t}) - e^{-3t}(-2e^{-2t}) \\[6pt] &= -3e^{-5t} + 2e^{-5t} \\[6pt] &= -e^{-5t}. \end{align}Example. Let $y_1 = e^{r_1 t}$, $y_2 = e^{r_2 t}$. Then
$$W(y_1, y_2)(t) = \begin{vmatrix} e^{r_1 t} & e^{r_2 t} \\ r_1 e^{r_1 t} & r_2 e^{r_2 t} \end{vmatrix} = (r_1 - r_2)e^{(r_2 + r_1)t}.$$Example. Let $y_1 = \cos t$, $y_2 = \sin t$. Then
$$W(y_1, y_2)(t) = \begin{vmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{vmatrix} = \cos^2 t + \sin^2 t = 1,$$for all $t$. So the Wronskian is a constant function.
Find the Wronskians of each of the following pairs and determine on which interval the two functions are linearly independent.
We deduce from the above theorem of existence and uniqueness the following.
Consider the DE
$$L(y) = y'' + p(t)y' + q(t)y = 0, \tag{1}$$where $p(t)$ and $q(t)$ are continuous on an open interval $I$ with initial condition
$$y(t_0) = y_0, \qquad y'(t_0) = y_0'.$$Suppose $y_1$, $y_2$ are two solutions to the DE (1). If
$$W(y_1, y_2)(t_0) = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \neq 0$$for $t_0$ in $I$, then there are $c_1$, $c_2$ such that the IVP is solved by
$$y(t) = c_1 y_1(t) + c_2 y_2(t).$$So the main point in the above theorem is that one can find $c_1$, $c_2$ to solve the IVP provided that the Wronskian $W(y_1, y_2)(t_0) \neq 0$.
Solving the two equations for $c_1$ and $c_2$ yields
\begin{align} y_1'(t_0)y_0 &= y_1'(t_0)y(t_0) = c_1 y_1'(t_0)y_1(t_0) + c_2 y_1'(t_0)y_2(t_0), \\[6pt] y_1(t_0)y_0' &= y_1(t_0)y'(t_0) = c_1 y_1(t_0)y_1'(t_0) + c_2 y_1(t_0)y_2'(t_0). \end{align}Subtracting the two equations yields
\begin{align} y_1'(t_0)y_0 - y_1(t_0)y_0' &= c_2[y_1'(t_0)y_2(t_0) - y_1(t_0)y_2'(t_0)] \\[6pt] &= -c_2 W(y_1, y_2)(t_0). \end{align}Since the $W(y_1, y_2)(t_0) \neq 0$, so we could solve for $c_2$ by
$$c_2 = \frac{y_1(t_0)y_0' - y_1'(t_0)y_0}{W(y_1, y_2)(t_0)} = \frac{W(y_1, y)(t_0)}{W(y_1, y_2)(t_0)} = \frac{\begin{vmatrix} y_1(t_0) & y_0 \\ y_1'(t_0) & y_0' \end{vmatrix}}{\begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{vmatrix}}.$$Similarly,
$$c_1 = \frac{y_2'(t_0)y_0 - y_2(t_0)y_0'}{W(y_1, y_2)(t_0)} = \frac{W(y, y_2)(t_0)}{W(y_1, y_2)(t_0)} = \frac{\begin{vmatrix} y_0 & y_2(t_0) \\ y_0' & y_2'(t_0) \end{vmatrix}}{\begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{vmatrix}}.$$We may rephrase the above IVP theorem into the following statement.
Suppose $y_1$, $y_2$ are two solutions to the DE
$$L(y) = y'' + p(t)y' + q(t)y = 0,$$where $p(t)$ and $q(t)$ are continuous on certain interval $I$. If
$$W(y_1, y_2)(t_0) = y_1(t_0)y_2'(t_0) - y_1'(t_0)y_2(t_0) \neq 0$$for some $t_0$ in $I$, then every solution $\phi(t)$ of the DE can be written as
$$y(t) = c_1 y_1(t) + c_2 y_2(t)$$for some constants $c_1$ and $c_2$.
Let $y_1(t)$ and $y_2(t)$ be two solutions (functions) to the DE
$$L(y) = y'' + p(t)y' + q(t)y = 0,$$where $p(t)$ and $q(t)$ are continuous on an open interval $I$. If $W(y_1, y_2)(t_0) \neq 0$ for some point $t_0$ in $I$, then we call
Let $y_1$, $y_2$ be solutions to the DE
$$L(y) = y'' + p(t)y' + qy = 0,$$where $p(t)$ and $q(t)$ are continuous on an open interval $I$. Then the Wronskian
$$W(y_1, y_2)(t) = c \cdot \exp\Big[-\int p(s)\,ds\Big]$$for all $t$ in $I$, where $c$ is a constant that depends on $y_1$, $y_2$ but not on $t$. Moreover, either $W(y_1, y_2)(t) \neq 0$ for all $t$ in $I$ or $W(y_1, y_2)(t) \equiv 0$ for all $t$ in $I$.
We multiply the first Eqn by $-y_2$ and added to the second Eqn. multiply by $y_1$ yields
$$y_2 y_1'' - y_2'' y_1 + p(t)(y_2 y_1' - y_2' y_1) = 0.$$Let $W(t) = W(y_1, y_2)(t)$. Notice that
$$W'(t) = y_2 y_1'' - y_2'' y_1.$$So the $W$ satisfies the first order DE:
$$W' + p(t)W(t) = 0.$$Therefore this equation admits a solution of the form
$$W(y_1, y_2)(t) = c \cdot \exp\Big[-\int p(s)\,ds\Big],$$from elementary consideration and the constant $c$ is independent of $t$. This completes the derivation.
Let $y_1$ and $y_2$ be two functions over $I$. We define two functions $y_1$ and $y_2$
$W(y_1, y_2)(t) \neq 0$ over $I$, if and only if $y_1$ and $y_2$ are linearly independent over $I$.
and $c_1$, $c_2$ do not vanish simultaneously. We differentiate the equation once
$$c_1 y_1'(t) + c_2 y_2'(t) = 0, \quad \text{all } t \text{ in } I.$$We use the first equation to eliminate $c_1$ from the second equation. This yields
$$0 = -c_2(y_2/y_1) \cdot y_1' + c_2 y_2' = c_2 \frac{y_1 y_2' - y_2 y_1'}{y_1}.$$Since $y_1$ is not identically zero, so we must have
$$0 = c_2(y_1 y_2' - y_2 y_1') = c_2 W(y_1, y_2)(t).$$But $W(y_1, y_2)(t) \neq 0$ over $I$. Hence $c_2 = 0$. But then $c_1 = 0$. This contradicts the assumption that $c_1$, $c_2$ do not vanish simultaneously. Hence $y_1$, $y_2$ are linearly independent over $I$.
Conversely, suppose $y_1$, $y_2$ are linearly independent. Then we want to prove $W(y_1, y_2)(t) \neq 0$ over $I$. We could prove this statement if we prove the contrapositive statement instead. That is, if there is a $t_0$ in $I$ such that $W(y_1, y_2)(t_0) = 0$, then $y_1$, $y_2$ are linearly dependent. But Abel's formula implies that $W(y_1, y_2)(t) \equiv 0$. Hence
$$c_1 y_1(t) + c_2 y_2(t) = 0, \quad \text{all } t \text{ in } I,$$and
$$c_1 y_1'(t) + c_2 y_2'(t) = 0, \quad \text{all } t \text{ in } I,$$hold simultaneously. The same argument shows that
$$0 = c_2(y_1 y_2' - y_2 y_1') = c_2 W(y_1, y_2)(t).$$But this time $W(y_1, y_2)(t) \equiv 0$ so that $c_2$ can take arbitrary non-zero value. Hence $y_1$, $y_2$ are linearly dependent, as required.
This means that if $\{y_1, y_2\}$ is a fundamental set of solutions of the DE, then they must be linearly independent.
Example (revisited). Let $y_1 = e^{r_1 t}$, $y_2 = e^{r_2 t}$. Then
$$W(y_1, y_2)(t) = (r_1 - r_2)e^{(r_2 + r_1)t} \neq 0$$everywhere, provided that $r_1 \neq r_2$. Hence $\{e^{r_1 t}, e^{r_2 t}\}$ is a fundamental set of solutions to the DE
$$(D - r_1)(D - r_2)y = 0,$$where $D = \frac{d}{dt}$.
Example (revisited). Let $y_1 = \cos t$, $y_2 = \sin t$. Then
$$W(y_1, y_2)(t) = \cos^2 t + \sin^2 t = 1,$$for all $t$. So $\{\cos t, \sin t\}$ is a fundamental set of solutions to the DE
$$(D + i)(D - i)y = (D^2 + 1)y = y'' + y = 0,$$where $D = \frac{d}{dt}$.
Example (Linear dependence). Test if the $\{\sin t, \cos(t - \frac{\pi}{2})\}$ is a fundamental set for some DE.
It is easy to check that one has
$$W\big(\sin t, \cos(t - \tfrac{\pi}{2})\big) = 0$$for all $t$. Hence the $\{\sin t, \cos(t - \frac{\pi}{2})\}$ is not a fundamental set for any DE.
Example. Show that
$$2t^2 y'' + 3ty' - y = 0, \quad t > 0,$$admits a fundamental set of solutions $\{y = t^{1/2}, y_2 = t^{-1}\}$.
We leave the readers to verify that the $y_1$, $y_2$ are both solutions to the DE. It remains to verify
$$W(y_1, y_2)(t) = \begin{vmatrix} t^{1/2} & t^{-1} \\ \frac{1}{2}t^{-1/2} & -t^{-2} \end{vmatrix} = -t^{-3/2} - \frac{1}{2}t^{-3/2} = -\frac{3}{2}t^{-3/2} \neq 0$$provided $t > 0$ (or $t < 0$). So the $y_1$, $y_2$ is a fundamental set of solutions.
Example. We recall that the DE
$$ay'' + by' + cy = 0,$$admits two characteristic roots
$$r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \lambda \pm i\mu$$and $b^2 - 4ac < 0$, admits two solutions $e^{\lambda}\cos\mu$ and $e^{\lambda}\sin\mu$. We compute their Wronskian:
\begin{align} W(e^{\lambda t}\cos\mu t,\; e^{\lambda t}\sin\mu t) &= \begin{vmatrix} e^{\lambda t}\cos\mu t & e^{\lambda t}\sin\mu t \\ (e^{\lambda t}\cos\mu t)' & (e^{\lambda t}\sin\mu t)' \end{vmatrix} \\[6pt] &= \begin{vmatrix} e^{\lambda t}\cos\mu t & e^{\lambda t}\sin\mu t \\ e^{\lambda t}(\lambda\cos\mu t - \mu\sin\mu t) & e^{\lambda t}(\lambda\sin\mu t + \mu\cos\mu t) \end{vmatrix} \\[6pt] &= \mu e^{2\lambda t} \end{align}which is non-vanishing throughout the real axis. Hence $\{e^{r_1 t}, e^{r_2 t}\}$ forms a fundamental set of solutions to the DE even if the $r_1$, $r_2$ are complex (conjugates).
The following theorem guarantees that a given second order linear differential equation must have a fundamental set of solutions.
The DE
$$L(y) = y'' + p(t)y' + q(t)y = 0,$$where $p(t)$ and $q(t)$ are continuous on an open interval $I$ such that for an arbitrarily chosen $t_0$ in $I$ always admit a fundamental set of solutions $\{y_1, y_2\}$ the following initial conditions hold
\begin{align} y_1(t_0) = 0, \quad &y_1'(t_0) = 1, \\[6pt] y_2(t_0) = 1, \quad &y_2'(t_0) = 0. \end{align}admit a unique solution. Let them be $y_1$ and $y_2$ respectively. But then
$$W(y_1, y_2)(t) = \begin{vmatrix} y_1(t_0) & y_2(t_0) \\ y_1'(t_0) & y_2'(t_0) \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1.$$Example (B&D). Find the fundamental set of solutions $y_1$, $y_2$ of the DE
$$y'' - y = 0,$$that is described by the existence of fundamental set theorem.
It is a standard procedure to check that both of $y_1 = e^t$, $y_2 = e^{-t}$ forms a fundamental set (of solutions) to the DE: $W(e^t, e^{-t})(t) = -2 \neq 0$. The remark after the IVP theorem guarantees with the initial point $t_0 = 0$ that there are constants $c_1$, $c_2$ and $d_1$, $d_2$ so that the expected fundamental set
\begin{align} y_3(t_0) = 0, \quad &y_3'(t_0) = 1, \\[6pt] y_4(t_0) = 1, \quad &y_4'(t_0) = 0. \end{align}can be written in the form
\begin{align} y_3(t) &= c_1 e^t + c_2 e^{-t}, \\[6pt] y_4(t) &= d_1 e^t + d_2 e^{-t}. \end{align}Indeed, it is a simple exercise to verify that $c_1 = \frac{1}{2}$, $c_2 = \frac{1}{2}$ and $d_1 = \frac{1}{2}$, $d_2 = -\frac{1}{2}$. That is,
$$W(y_3, y_4)(t) = W(y_3, y_4)(t) = \cosh^2 t - \sinh^2 t = 1.$$ $$W(\cosh t, \sinh t) = \begin{vmatrix} \cosh t & \sinh t \\ \sinh t & \cosh t \end{vmatrix} = \cosh^2 t - \sinh^2 t = 1.$$So the $\{\cosh t, \sinh t\}$ forms a fundamental set of solutions that meets the question asked.
Example (Revisited). Recall that $y_1(t) = e^{-2t}$, $y_2(t) = e^{-3t}$ are two solutions of the DE
$$y'' + 5y' + 6 = 0$$with
$$W(f, g)(t) = -e^{-5t} \neq 0.$$Thus the $y_1(t) = e^{-2t}$, $y_2(t) = e^{-3t}$ forms a fundamental set of solutions to differential equation. So every solution $y$ of the DE on the real-axis can be written in the form $y(t) = c_1 e^{-2t} + c_2 e^{-3t}$ for some constants $c_1$, $c_2$.
Let $\{y_1, y_2\}$ be a fundamental set of a second order ODE. Let $y_3 = a_1 y_1 + a_2 y_2$ and $y_4 = a_3 y_1 + a_4 y_2$. Show
$$W(y_3, y_4) = (a_1 b_2 - a_2 b_1)\,W(y_1, y_2).$$Discuss under what condition that $\{y_3, y_4\}$ can be a fundamental set of the same second order ODE.
Let $y_1 = e^{2t}$ and $y_2 = e^{-3t}$. Compute the Wronskian $W(y_1, y_2)(t)$.
The Wronskian $W(f, g)$ is defined as:
What is $y_1' = \frac{d}{dt}(e^{2t})$?
What is $y_2' = \frac{d}{dt}(e^{-3t})$?
$W = y_1 y_2' - y_1' y_2 = e^{2t}(-3e^{-3t}) - (2e^{2t})(e^{-3t}) = ?$
Since $W(y_1, y_2) = -5e^{-t} \neq 0$ for all $t$, what can we conclude?
Consider the DE $y'' + 4y' + 3y = 0$. If $y_1, y_2$ are solutions with $W(y_1, y_2)(0) = 2$, find $W(y_1, y_2)(t)$ using Abel's formula.
For $y'' + p(t)y' + q(t)y = 0$, Abel's formula states:
In the DE $y'' + 4y' + 3y = 0$, what is $p(t)$?
What is $\displaystyle\int p(t)\,dt = \int 4\,dt$?
By Abel's formula, $W(t) = c \cdot e^{-4t}$. Using $W(0) = 2$, find $c$:
Therefore, $W(y_1, y_2)(t) = ?$
Determine whether $y_1 = \sin t$ and $y_2 = \cos(t - \frac{\pi}{2})$ form a fundamental set of solutions.
Using the identity $\cos(\theta - \frac{\pi}{2}) = \sin\theta$, what is $y_2$?
If $y_1 = \sin t$ and $y_2 = \sin t$, what is $W(y_1, y_2)$?
Since $W(y_1, y_2) \equiv 0$, what can we conclude?
Do $\{\sin t, \cos(t - \frac{\pi}{2})\}$ form a fundamental set?
— End of Wronskians and Independence Notes —