17.2: First Order Homogeneous Linear Equations

A simple, but important and useful, type of separable equation is the first order homogeneous linear equation:

Definition: first order homogeneous linear differential equation

A first order homogeneous linear differential equation is one of the form

$$\dot y + p(t)y=0 \nonumber$$

or equivalently

$$\dot y = -p(t)y. \nonumber$$

"Linear'' in this definition indicates that both $\dot y$ and $y$ occur to the first power; "homogeneous'' refers to the zero on the right hand side of the first form of the equation.

Example $\PageIndex{2}$

The equation $\dot y = 2t(25-y)$ can be written $\dot y + 2ty= 50t$. This is linear, but not homogeneous. The equation $\dot y=ky$, or $\dot y-ky=0$ is linear and homogeneous, with a particularly simple $p(t)=-k$.

Because first order homogeneous linear equations are separable, we can solve them in the usual way:

$$\eqalign{ \dot y &= -p(t)y\cr \int {1\over y}\,dy &= \int -p(t)\,dt\cr \ln|y| &= P(t)+C\cr y&=\pm\,e^{P(t)}\cr y&=Ae^{P(t)},\cr} \nonumber$$

where $P(t)$ is an anti-derivative of $-p(t)$. As in previous examples, if we allow $A=0$ we get the constant solution $y=0$.

Example $\PageIndex{3}$

Solve the initial value problems $\dot y + y\cos t =0$, $y(0)=1/2$ and $y(2)=1/2$.

Solution

We start with

$$P(t)=\int -\cos t\,dt = -\sin t, \nonumber$$

so the general solution to the differential equation is

$$y=Ae^{-\sin t}. \nonumber$$

To compute $A$ we substitute:

$${1\over 2} = Ae^{-\sin 0} = A, \nonumber$$

so the solutions is

$$y = {1\over 2} e^{-\sin t}. \nonumber$$

For the second problem,

$$\eqalign{{1\over 2} &= Ae^{-\sin 2}\cr A &= {1\over 2}e^{\sin 2}\cr} \nonumber$$

so the solution is

$$y = {1\over 2}e^{\sin 2}e^{-\sin t}. \nonumber$$

Example $\PageIndex{4}$

Solve the initial value problem $y\dot y+3y=0$, $y(1)=2$, assuming $t>0$.

Solution

We write the equation in standard form: $\dot y+3y/t=0$. Then

$$P(t)=\int -{3\over t}\,dt=-3\ln t \nonumber$$

and

$$y=Ae^{-3\ln t}=At^{-3}. \nonumber$$

Substituting to find $A$: $2=A(1)^{-3}=A$, so the solution is $y=2t^{-3}$.

Contributors

• David Guichard (Whitman College)

• Integrated by Justin Marshall.