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# 17.2: First Order Homogeneous Linear Equations

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

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

or equivalently

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

"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 $$\PageIndex{2}$$

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}

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

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

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 $$\PageIndex{4}$$

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

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

and

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

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

## Contributors

• Integrated by Justin Marshall.