### Differential equations: Linear first-order differential equations

### Linear first-order ODE and integrating factor

We begin by showing how to solve linear first-order ODEs with constant coefficients.

Solution of first-order linear ODE with constant coefficients

The general solution of the first order linear ODE \[ \dfrac{\dd y}{\dd t}+p\cdot y=q\cdot t\] in the unknown function #y# of #t#, where #p# and #q# are constants with #p\ne0#, is

\[y(t)=C\cdot\e^{-p\cdot t}+\frac{q}{p}\cdot t-\frac{q}{p^2}\]

Here, #C# is the integration constant.

If #p=0#, the general solution is the integral #y = \int q\cdot t\,\dd t = \frac{1}{2}q\cdot t^2+C#.

In the proof above, the differential equation is multiplied by a function that makes it possible to bring the terms with #y# and #\dd y# under one differential. We will deal with the technique in greater detail.

Integrating factor

In the proof of the statement we first wrote the ODE in differential form and next multiplied all the terms by #\e^{p\cdot t}#. That factor enabled us to collect the terms with #{\dd y}# and #y\,\dd t# into one differential (that is, under one #\dd#). Such a factor is an **integrating factor**.

In the case of a homogeneous linear first order ODE #y'+p(t)\cdot y = 0#, the function #\e^{P(t)}#, where #P(t)# is an antiderivative of #p(t)#, is an integrating factor.

Below are some examples.

Specify the general solution of this equation.

Previously, we solved this ODE by means of separation of variables.

The theorem

*Solution of first-order linear ODE with constant coefficients*gives that the solution is \[y(t)=C\cdot\e^{-p\cdot t}+\frac{q}{p}\cdot t-\frac{q}{p^2}\] where #p=-r# and #q=0#. This immediately leads to the answer \(y = C\cdot\e^{r\cdot t}\).

As a third alternative solution strategy, we show how to use an integrating factor in order to find the answer:

\[\begin{array}{rcl}\dd y-r\cdot y\,\dd t&=&0\\

&&\phantom{x}\color{blue}{\text{differential form}}\\

\e^{-r\cdot t }\dd y-\e^{-r\cdot t }\cdot r\cdot y\,\dd t&=&0\\

&&\phantom{x}\color{blue}{\text{differential form multiplied by integrating factor;}}\\

&&\phantom{x}\color{blue}{\text{this factor has the form }\e^{-r\cdot t}}\\

&&\phantom{x}\color{blue}{\text{since }{-r\cdot t}\text{ is an antiderivative of }-r}\\

&&\phantom{x}\color{blue}{\text{so the left side can be written as follows}}\\

\dd\left(\e^{-r\cdot t } \cdot y\right)&=&0\\

&&\phantom{x}\color{blue}{\text{differential form under a single differential}}\\

\e^{-r\cdot t }\cdot y&=&C\\

&&\phantom{x}\color{blue}{\text{differential form integrated with integration constant }C}\\

y&=&C\cdot\e^{r\cdot t } \\

&&\phantom{x}\color{blue}{\text{ solved for }y}\\

\end{array}\]

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