First Order Linear Differential Equations

Subjects: Ordinary Differential Equations
Links: First Order Differential Equations

Def: A first order linear differential equation is of the form

y^{'} + p (t) y = q (t)

where $p$ and $q$ are given functions of the independent variable $t$ . This equation is linear in $y$ , that is why is called linear.

Existence and Uniqueness

If the functions $p$ and $g$ are continuous on an open interval $I = (α, β)$ containing the point $t = t_{0}$ , then there exists a unique function $y = ϕ (t)$ that satisfies the differential equation

y^{'} + p (t) y = g (t)

for each $t$ in $I$ , and that also satisfies the initial condition

y (t_{0}) = y_{0},

where $y_{0}$ is an arbitrary prescribed initial value.

Method

The most effective method to solve this kind of equation is by using an integrating factor. The main thing we want to see is left-hand side of the equation is of of the form of a product. We multiply by an unknown function $μ (t)$ so we have the equation

μ (t) y^{'} + μ (t) p (t) y = μ (t) q (t)

We make the left-hand side of the equation equal to the derivative of $μ (t) y$ , then we get that

\frac{d}{d t} (μ (t) y) = μ (t) y^{'} + μ^{'} (t) y = μ (t) y^{'} + μ (t) p (t) y

If we simplify we get that

μ^{'} (t) = μ (t) p (t)

this is a special kind of differential equation solved by having $μ (t) = c_{1} \exp (\int p (t) d t)$ with $c_{1} \neq 0$ , where $\int p (t) d t$ represents the the anti derivative of $p (t)$ . Plugging it back to the original equation we get that

\frac{d}{d t} (μ (t) y) = μ (t) q (t)

Antiderivating we get that

μ (t) y = \int μ (t) g (t) d t + C

if we solve for $y$ we get that

y (t) = \frac{1}{μ (t)} \int μ (t) g (t) d t + \frac{C}{μ (t)}

The value of $C$ is through the initial conditions, and this is final equation shows that the choice of $c_{1}$ doesn’t matter since it will get canceled as long as $c_{1} \neq 0$ .