Existence/Uniqueness of Solutions to First Order Linear Differential Eqs.

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

Suppose that we have a linear differential equation $y' + p(t) y = g(t)$ and that we want to solve the initial value problem of $y(t_0) = y_0$. We have already looked at various methods to solve these sort of linear differential equations, however, we will now ask the question of whether or not solutions exist and whether or not these solutions are unique. The following theorem will provide sufficient conditions allowing the unique existence of a solution to these initial value problems.

 Theorem 1: Let $p$ and $g$ be continuous functions on the open interval $I = ( \alpha, \beta)$, and let $t_0 \in (\alpha, \beta)$. Then for each $t \in I$ there exists a unique solution $y = \phi (t)$ to the differential equation $\frac{dy}{dt} + p(t) y = g(t)$ that also satisfies the initial value condition that $y(t_0) = y_0$.
• Proof: Let $p = p(t)$ and $g = g(t)$ be continuous on $I = (\alpha, \beta)$ and let $t_0 \in (\alpha, \beta)$. Suppose that we have the differential equation $\frac{dy}{dt} + p(t) y = g(t)$. Let $\mu (t)$ be such that $\mu ' (t) = \mu (t) p(t)$ (i.e, $\mu (t)$ is an integrating factor of this differential equation).
• Since $p(t)$ is a continuous function in $I = (\alpha, \beta)$ we have that the integrating factor $\mu (t) = e^{\int p(t) \: dt}$ is defined on $I = (\alpha, \beta)$ and is also nonzero and differentiable (sinice $\int p(t) \: dt$ exists from the continuity of $p(t)$ and $\mu (t) = e^{\int p(t) \: dt} > 0$ as an exponential function). Multiplying the differential equation above by $\mu (t)$ and we have that:
(1)
\begin{align} \quad \frac{dy}{dt} + p(t) = g(t) \\ \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{d}{dt} \left ( \mu (t) y \right ) = \mu (t) g(t) \\ \end{align}
• Note since both $\mu (t)$ and $g(t)$ are continuous on $I = (\alpha, \beta)$ we have that their product $\mu (t) g(t)$ is also continuous on $I = (\alpha, \beta)$. Therefore $\mu (t) g(t)$ is integrable. Integrating both sides of the equation above and we have that:
(2)
\begin{align} \quad \int \frac{d}{dt} \left ( \mu (t) y \right ) \: dt = \int \mu (t) g(t) \: dt \\ \quad \mu (t) y = \int \mu (t) g(t) \: dt + C \end{align}
• Once again, since $\mu (t) \neq 0$ we can divide both sides of the equation above by $\mu (t)$ which isolates $y$ and proves the existence of a solution to the original differential equation.
(3)
\begin{align} \quad y = \frac{1}{\mu (t)} \int \mu (t) g(t) \: dt + C \end{align}
• Now we only need to show the uniqueness of a solution given the initial condition $y(t_0) = y_0$ for $t_0 \in I$. Now since $y(t_0) = y_0$ is the initial condition of our differential equation, then we know that this determines the value of the constant $C$ and the solution is unique. Let the integrating factor $\mu (t) = e^{\int_{t_0}^{t} p(s) \: ds}$ and let $C = y_0$. Then
(4)
\begin{align} \quad y = \frac{1}{\mu (t)} \int_{t_0}^{t} \mu (s) g(s) \: ds + y_0 = \frac{1}{e^{\int_{t_0}^{t} p(s) \: ds}} \int_{t_0}^{t} \mu (s) g(s) \: ds + y_0 \end{align}
• Notice that $\mu (t_0) = e^{\int_{t_0}^{t_0} p(s) \: ds} = e^0 = 1$. So $y(t_0)$ is:
(5)
\begin{align} \quad y(t_0) = \frac{1}{e^{\int_{t_0}^{t_0} p(s) \: ds} } \int_{t_0}^{t_0} \mu (s) g(s) \: ds + y_0 = 1 \cdot 0 + y_0 = y_0 \quad \blacksquare \end{align}

## Example 1

Find the interval for which the linear differential equation $\frac{dy}{dt} + 4y = \cos t$ has a unique solution for the initial condition $y(0) = 0$ and then find this solution.

Note that in this example $p(t) = 4$ and $g(t) = \cos t$. Both of these functions are continuous on all of $\mathbb{R}$, and so the unique solution is given on the interval $(-\infty, \infty)$.

We will now solve for this solution. Let $\mu (t) = e^{\int 4 \: dt} = e^{4t}$ be an integrating factor for our differential equation. Multiplying both sides of the differential equation above by $\mu (t)$ and we have that:

(6)
\begin{align} \quad e^{4t} \frac{dy}{dt} + 4e^{4t} y = e^{4t} \cos t \\ \quad \frac{d}{dt} \left ( e^{4t} y \right ) = e^{4t} \cos t \\ \quad e^{4t} y = \int e^{4t} \cos t \: dt \end{align}

We want to evaluate $\int e^{4t} \cos t \: dt$. We can do this through Integration by Parts or using Tabular Integration. Either way, we get that $\int e^{4t} \cos t \: dt = \frac{e^{4t} ( \sin t + 4 \cos t)}{17} + C$. Hence:

(7)
\begin{align} \quad e^{4t} y = \frac{e^{4t} ( \sin t + 4 \cos t)}{17} + C \end{align}

Now since $y(0) = 0$ we have that $C = -\frac{4}{17}$, so our unique solution to the given differential equation is:

(8)
\begin{align} \quad e^{4t} y = \frac{e^{4t} (\sin t + 4 \cos t)}{17} - \frac{4}{17} \\ \quad y = \frac{\sin t + 4 \cos t}{17} - \frac{4}{17e^{4t}} \end{align}