$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 17.1: First Order Differential Equations

• • Contributed by David Guichard
• Professor (Mathematics) at Whitman College

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

We start by considering equations in which only the first derivative of the function appears.

Definition 17.1.1: First Order Differential Equation

A first order differential equation is an equation of the form $$F(t, y, \dot{y})=0$$.

A solution of a first order differential equation is a function $$f(t)$$ that makes $$F(t,f(t),f'(t))=0$$ for every value of $$t$$.

Here, $$F$$ is a function of three variables which we label $$t$$, $$y$$, and $$\dot{y}$$. It is understood that $$\dot{y}$$ will explicitly appear in the equation although $$t$$ and $$y$$ need not. The term "first order'' means that the first derivative of $$y$$ appears, but no higher order derivatives do.

Example $$\PageIndex{2}$$:

The equation from Newton's law of cooling, $$\dot{y}=k(M-y)$$ is a first order differential equation; $$F(t,y,\dot y)=k(M-y)-\dot y$$.

Example $$\PageIndex{3}$$:

$$\dot{y}=t^2+1$$ is a first order differential equation; $$F(t,y,\dot y)= \dot y-t^2-1$$. All solutions to this equation are of the form $$t^3/3+t+C$$.

Definition 17.1.4: First Order Initial Value Problem

A first order initial value problem is a system of equations of the form $$F(t, y, \dot{y})=0$$, $$y(t_0)=y_0$$. Here $$t_0$$ is a fixed time and $$y_0$$ is a number.

A solution of an initial value problem is a solution $$f(t)$$ of the differential equation that also satisfies the initial condition $$f(t_0) = y_0$$.

Example $$\PageIndex{5}$$:

The initial value problem $$\dot{y}=t^2+1$$, $$y(1)=4$$ has solution $$f(t)=t^3/3+t+8/3$$.

The general first order equation is rather too general, that is, we can't describe methods that will work on them all, or even a large portion of them. We can make progress with specific kinds of first order differential equations. For example, much can be said about equations of the form $$\dot{y} = \phi (t, y)$$ where $$\phi$$ is a function of the two variables $$t$$ and $$y$$. Under reasonable conditions on $$\phi$$, such an equation has a solution and the corresponding initial value problem has a unique solution. However, in general, these equations can be very difficult or impossible to solve explicitly.

Example $$\PageIndex{6}$$:

Consider this specific example of an initial value problem for Newton's law of cooling: $$\dot y = 2(25-y)$$, $$y(0)=40$$. We first note that if $$y(t_0) = 25$$, the right hand side of the differential equation is zero, and so the constant function $$y(t)=25$$ is a solution to the differential equation. It is not a solution to the initial value problem, since $$y(0)\not=40$$. (The physical interpretation of this constant solution is that if a liquid is at the same temperatureas its surroundings, then the liquid will stay at that temperature.) So long as $$y$$ is not 25, we can rewrite the differential equation as

\eqalign{{dy\over dt}{1\over 25-y}&=2\cr{1\over 25-y}\,dy&=2\,dt,\cr}

so

$$\int {1\over 25-y}\,dy = \int 2\,dt,$$

that is, the two anti-derivatives must be the same except for a constant difference. We can calculate these anti-derivatives and rearrange the results:

\eqalign{\int {1\over 25-y}\,dy &= \int 2\,dt\cr (-1)\ln|25-y| &= 2t+C_0\cr \ln|25-y| &= -2t - C_0 = -2t + C\cr |25-y| &= e^{-2t+C}=e^{-2t} e^C\cr y-25 & = \pm\, e^C e^{-2t} \cr y &= 25 \pm e^C e^{-2t} =25+Ae^{-2t}.\cr}

Here $$A = \pm\, e^C = \pm\, e^{-C_0}$$ is some non-zero constant. Since we want $$y(0)=40$$, we substitute and solve for $$A$$:

\eqalign{40&=25+Ae^0\cr 15&=A,\cr}

and so $$y=25+15 e^{-2t}$$ is a solution to the initial value problem. Note that $$y$$ is never 25, so this makes sense for all values of $$t$$. However, if we allow $$A=0$$ we get the solution $$y=25$$ to the differential equation, which would be the solution to the initial value problem if we were to require $$y(0)=25$$. Thus, $$y=25+Ae^{-2t}$$ describes all solutions to the differential equation $$\dot y = 2(25-y)$$, and all solutions to the associated initial value problems.

Why could we solve this problem? Our solution depended on rewriting the equation so that all instances of $$y$$ were on one side of the equation and all instances of $$t$$ were on the other; of course, in this case the only $$t$$ was originally hidden, since we didn't write $$dy/dt$$ in the original equation. This is not required, however.

Example $$\PageIndex{7}$$:

Solve the differential equation $$\dot y = 2t(25-y)$$. This is almost identical to the previous example. As before, $$y(t)=25$$ is a solution. If $$y\not=25$$,

\eqalign{\int {1\over 25-y}\,dy &= \int 2t\,dt\cr(-1)\ln|25-y| &= t^2+C_0\cr \ln|25-y| &= -t^2 - C_0 = -t^2 + C\cr |25-y| &= e^{-t^2+C}=e^{-t^2} e^C\cr y-25 & = \pm\, e^C e^{-t^2} \cr y &= 25 \pm e^C e^{-t^2} =25+Ae^{-t^2}.\cr}

As before, all solutions are represented by $$y=25+Ae^{-t^2}$$, allowing $$A$$ to be zero.

Definition 17.1.8: Separable Differential Equation

A first order differential equation is separable if it can be written in the form $$\dot{y} = f(t) g(y)$$.

As in the examples, we can attempt to solve a separable equation by converting to the form

$$\int {1\over g(y)}\,dy=\int f(t)\,dt.$$

This technique is called separation of variables. The simplest (in principle) sort of separable equation is one in which $$g(y)=1$$, in which case we attempt to solve

$$\int 1\,dy=\int f(t)\,dt.$$

We can do this if we can find an anti-derivative of $$f(t)$$.

Also as we have seen so far, a differential equation typically has an infinite number of solutions. Ideally, but certainly not always, a corresponding initial value problem will have just one solution. A solution in which there are no unknown constants remaining is called a particular solution.

The general approach to separable equations is this: Suppose we wish to solve $$\dot{y} =f(t) g(y)$$ where $$f$$ and $$g$$ are continuous functions. If $$g(a)=0$$ for some $$a$$ then $$y(t)=a$$ is a constant solution of the equation, since in this case $$\dot y = 0 = f(t)g(a)$$. For example, $$\dot{y} =y^2 -1$$ has constant solutions $$y(t)=1$$ and $$y(t)=-1$$.

To find the nonconstant solutions, we note that the function $$1/g(y)$$ is continuous where $$g\not=0$$, so $$1/g$$ has an antiderivative $$G$$. Let $$F$$ be an antiderivative of $$f$$. Now we write

$$G(y) = \int {1\over g(y)}\,dy = \int f(t)\,dt=F(t)+C,$$

so $$G(y)=F(t)+C$$. Now we solve this equation for $$y$$.

Of course, there are a few places this ideal description could go wrong: we need to be able to find the antiderivatives $$G$$ and $$F$$, and we need to solve the final equation for $$y$$. The upshot is that the solutions to the original differential equation are the constant solutions, if any, and all functions $$y$$ that satisfy $$G(y)=F(t)+C$$.

Example $$\PageIndex{9}$$:

Consider the differential equation $$\dot y=ky$$. When $$k>0$$, this describes certain simple cases of population growth: it says that the change in the population $$y$$ is proportional to the population. The underlying assumption is that each organism in the current population reproduces at a fixed rate, so the larger the population the more new organisms are produced. While this is too simple to model most real populations, it is useful in some cases over a limited time. When $$k<0$$, the differential equation describes a quantity that decreases in proportion to the current value; this can be used to model radioactive decay.

The constant solution is $$y(t)=0$$; of course this will not be the solution to any interesting initial value problem. For the non-constant solutions, we proceed much as before:

\eqalign{\int {1\over y}\,dy&=\int k\,dt\cr \ln|y| &= kt+C\cr |y| &= e^{kt} e^C\cr y &= \pm \,e^C e^{kt} \cr y&= Ae^{kt}.\cr}

Again, if we allow $$A=0$$ this includes the constant solution, and we can simply say that $$y=Ae^{kt}$$ is the general solution. With an initial value we can easily solve for $$A$$ to get the solution of the initial value problem. In particular, if the initial value is given for time $$t=0$$, $$y(0)=y_0$$, then $$A=y_0$$ and the solution is $$y= y_0 e^{kt}$$.

## Contributors

• Integrated by Justin Marshall.