# The Existence/Uniqueness of Solutions to Second Order Linear Differential Equations

Recall that from The Existence/Uniqueness of Solutions to First Order Linear Differential Equations page that if $p$ and $g$ are continuous functions on an interval $I = (\alpha, \beta)$ and $t_0 \in I$, then the linear first order differential equations in the form $\frac{dy}{dt} + p(t) y = g(t)$ with the initial value condition $y(t_0) = y_0$ has a unique solution $y = \phi(t)$ for $t \in I$. We will now state an analogous theorem for linear second order differential equations.

Theorem 1: Let $p$, $q$, and $g$ be continuous functions on an open interval $I$ such that $t_0 \in I$. Then the second order linear differential equation $\frac{d^2 y}{dt^2} + p(t) \frac{dy}{dt} + q(t) y = g(t)$ with the initial conditions $y(t_0) = y_0$ and $y'(t_0) = y’_0$ has a unique solution $y = \phi (t)$ throughout $I$. |

We will now look at an example of applying Theorem 1.

## Example 1

**Determine the largest interval for which the second order linear differential equation $(t^2 – 4t) \frac{d^2y}{dt^2} + 3t \frac{dy}{dt} + 4y = 2$ with the initial conditions $y(3) = 0$ and $y'(3) = -1$ has a unique solution.**

In order to apply Theorem 1, we will need to rewrite our differential equation as follows:

(1)

Therefore $p(t) = \frac{3t}{t(t – 4)}$, $q(t) = \frac{4}{t(t-4)}$, and $g(t) = \frac{2}{t(t-4)}$. Collectively, it is not hard to see that all of these functions are continuous for all $t$ where their denominators are nonzero, that is for $t \neq 0$ and for $t \neq 4$. We note that our interval must contain our initial condition $y(3) = 0$, that is, the value $t = 3$ must be contained in our interval. Therefore a unique solution $y = \phi(t)$ to this second order linear differential equation exists on the interval $(0, 4)$.

## Example 2

**Determine the largest interval for which the second order linear differential equation $(t – 3) \frac{d^2y}{dt^2} + t \frac{dy}{dt} + \ln \mid t \mid y = 0$ with the initial conditions $y(1) = 0$ and $y'(1) = 1$ has a unique solution.**

In order to apply Theorem 1, we will need to rewrite our differential equation as follows.

(2)

Therefore $p(t) = \frac{t}{t -3}$, $q(t) = \frac{\ln \mid t \mid}{t – 3}$, and $g(t) = 0$. We have that $p(t)$ is continuous for $t \neq 3$, $q(t)$ is continuous for $t \ neq 3$ and $t \neq 0$, and $g(t)$ is continuous everywhere.

The possible intervals for the solution are therefore $(-\infty, 0)$, $(0, 3)$, or $(3, \infty)$.

With the initial condition $y(1) = 0$ we see that $1$ must be contained in our choice of interval, and so $(0, 3)$ is the largest interval for which a unique solution to our differential equation exists.

Remark 1: In both of the examples above, we needed the second order differential equation to be in the appropriate form $\frac{d^2y}{dt^2} + p(t) \frac{dy}{dt} + q(t)y = g(t)$ in order to apply Theorem 1. If we were to apply Theorem 1 without the second order differential equations from above in the correct form, then we would not obtain correct intervals for which a unique solution is guaranteed to be in. |