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

Definition 19.2.1 A first order homogeneous linear differential equation is one of the form $\ds \dot y + p(t)y=0$ or equivalently $\ds \dot y = -p(t)y$.

"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 19.2.2 The equation $\ds \dot y = 2t(25-y)$ can be written $\ds \dot y + 2ty= 50t$. This is linear, but not homogeneous. The equation $\ds \dot y=ky$, or $\ds \dot y-ky=0$ 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: $$\eqalign{ \dot y &= -p(t)y\cr \int {1\over y}\,dy = \int -p(t)\,dt\cr \ln|y| &= P(t)+C\cr y&=\pm\,e^{P(t)}\cr y&=Ae^{P(t)},\cr} $$ 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 19.2.3 Solve the initial value problems $\ds \dot y + y\cos t =0$, $y(0)=1/2$ and $y(2)=1/2$. We start with $$P(t)=\int -\cos t\,dt = -\sin t,$$ so the general solution to the differential equation is $$y=Ae^{-\sin t}.$$ To compute $A$ we substitute: $$ {1\over 2} = Ae^{-\sin 0} = A,$$ so the solutions is $$ y = {1\over 2} e^{-\sin t}.$$ For the second problem, $$ \eqalign{ {1\over 2} &= Ae^{-\sin 2}\cr A &= {1\over 2}e^{\sin 2}\cr} $$ so the solution is $$ y = {1\over 2}e^{\sin 2}e^{-\sin t}.$$

Example 19.2.4 Solve the initial value problem $y\dot y+3y=0$, $y(1)=2$, assuming $t>0$. We write the equation in standard form: $\dot y+3y/t=0$. Then $$P(t)=\int -{3\over t}\,dt=-3\ln t$$ and $$ y=Ae^{-3\ln t}=At^{-3}.$$ Substituting to find $A$: $\ds 2=A(1)^{-3}=A$, so the solution is $\ds y=2t^{-3}$.

Exercises 19.2

Find the general solution of each equation in 1–4.

Ex 19.2.1 $\ds\dot y+5y=0$ (answer)

Ex 19.2.2 $\ds\dot y-2y=0$ (answer)

Ex 19.2.3 $\ds\dot y+{y\over 1+t^2}=0$ (answer)

Ex 19.2.4 $\ds\dot y+t^2y=0$ (answer)

In 5–14, solve the initial value problem.

Ex 19.2.5 $\ds\dot y + y=0$, $y(0)=4$ (answer)

Ex 19.2.6 $\ds\dot y -3y=0$, $y(1)=-2$ (answer)

Ex 19.2.7 $\ds\dot y + y\sin t = 0$, $y(\pi)=1$ (answer)

Ex 19.2.8 $\ds\dot y +ye^t=0$, $y(0)=e$ (answer)

Ex 19.2.9 $\ds\dot y +y\sqrt{1+t^4}=0$, $y(0)=0$ (answer)

Ex 19.2.10 $\ds\dot y + y\cos(e^t)=0$, $y(0)=0$ (answer)

Ex 19.2.11 $\ds t\dot y - 2y = 0$, $y(1)=4$ (answer)

Ex 19.2.12 $\ds t^2\dot y + y = 0$, $y(1)=-2$, $t>0$ (answer)

Ex 19.2.13 $\ds t^3\dot y = 2y$, $y(1)=1$, $t>0$ (answer)

Ex 19.2.14 $\ds t^3\dot y = 2y$, $y(1)=0$, $t>0$ (answer)

Ex 19.2.15 A function $y(t)$ is a solution of $\ds\dot y + ky=0$. Suppose that $y(0)=100$ and $y(2)=4$. Find $k$ and find $y(t)$. (answer)

Ex 19.2.16 A function $y(t)$ is a solution of $\ds\dot y + t^ky=0$. Suppose that $y(0)=1$ and $y(1)=e^{-13}$. Find $k$ and find $y(t)$. (answer)

Ex 19.2.17 A bacterial culture grows at a rate proportional to its population. If the population is one million at $t=0$ and 1.5 million at $t=1$ hour, find the population as a function of time. (answer)

Ex 19.2.18 A radioactive element decays with a half-life of 6 years. If a mass of the element weighs ten pounds at $t=0$, find the amount of the element at time $t$. (answer)