A non-homogeneous first-order linear differential equation

Ordinary differential equations: Solving ODEs by an integrating factor

A non-homogeneous first-order linear differential equation

We consider the non-homogeneous first-order linear differential equation

$y'(t)=a(t)\cdot y(t)+b(t)$ where $a(t)\neq 0$ and $b(t)\neq 0$ . This means that the ODE

has order 1, because only the first derivative $y'(t)$ is present;
is linear because of the linear form in which $y$ and $y'$ are present;
non-homogeneous as $b(t)\neq 0$ .

Let $A(t)$ be an antiderivative of $a(t)$ (it exists when $a(t)$ is continuous on a closed interval).
Then, $e^{-A(t)}$ is an integrating factor.

Let $a(t)$ and $b(t)$ be nonzero continuous function and let $A(t)$ be an antiderivative of $a(t)$ . Then the general solution of the ODE

$y'(t)=a(t)\cdot y(t)+b(t)$ is

$y(t)=e^{A(t)}\cdot (F(t)+c)$ where $F(t)$ is an antiderivative of $e^{-A(t)}\cdot b(t)$ and $c$ is an integration constant.

We first write the differential equation in the form

$y'(t) -a(t)\cdot y(t)=b(t)$ Multiplication of the left- and right-hand side of the equation by the integrating factor $e^{-A(t)}$ gives:

$e^{-A(t)}\cdot y'(t) - e^{-A(t)}\cdot a(t)\cdot y(t) = e^{-A(t)}\cdot b(t)$ The left-hand side of the equation can be simplified by using the product rule and the chain rule for differentiation:

$\left( e^{-A(t)}\cdot y(t)\right)' =e^{-A(t)}\cdot b(t)$ Let $F(t)$ be an antiderivative of $e^{-A(t)}\cdot b(t)$ , then we get by integration:

$e^{-A(t)}\cdot y(t) = F(t) + c$ where $c$ is an integration constant. Rewriting leads to the general solution

$y(t)=e^{A(t)}\cdot (F(t)+c)$ The general solution of the ODE is equal to the sum of the specific solution $\displaystyle y=e^{A(t)}\cdot F(t)$ and the general solution $y=c\cdot e^{A(t)}$ of the homogeneous equation $\displaystyle\frac{dy}{dt}=a(t)\cdot y\tiny.$

The choice of $c$ in the ODE of the solution depends on the initial value $y(t_0)=y_0$ :

$c= e^{-A(t_0)}\cdot y_0 - F(t_0)$