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 \(y(t_0)=y_0\) : \[c= e^{-A(t_0)}\cdot y_0 - F(t_0)\]