17.2: First Order Homogeneous Linear Equations
( \newcommand{\kernel}{\mathrm{null}\,}\)
A simple, but important and useful, type of separable equation is the first order homogeneous linear equation:
A first order homogeneous linear differential equation is one of the form
˙y+p(t)y=0
or equivalently
˙y=−p(t)y.
"Linear'' in this definition indicates that both ˙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.
The equation ˙y=2t(25−y) can be written ˙y+2ty=50t. This is linear, but not homogeneous. The equation ˙y=ky, or ˙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:
˙y=−p(t)y∫1ydy=∫−p(t)dtln|y|=P(t)+Cy=±eP(t)y=AeP(t),
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.
Solve the initial value problems ˙y+ycost=0, y(0)=1/2 and y(2)=1/2.
Solution
We start with
P(t)=∫−costdt=−sint,
so the general solution to the differential equation is
y=Ae−sint.
To compute A we substitute:
12=Ae−sin0=A,
so the solutions is
y=12e−sint.
For the second problem,
12=Ae−sin2A=12esin2
so the solution is
y=12esin2e−sint.
Solve the initial value problem y˙y+3y=0, y(1)=2, assuming t>0.
Solution
We write the equation in standard form: ˙y+3y/t=0. Then
P(t)=∫−3tdt=−3lnt
and
y=Ae−3lnt=At−3.
Substituting to find A: 2=A(1)−3=A, so the solution is y=2t−3.
Contributors
Integrated by Justin Marshall.