17.2: First Order Homogeneous Linear Equations
- Page ID
- 4842
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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
\[\dot y + p(t)y=0 \nonumber \]
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.
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 \(y=0\).
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
\[ y = {1\over 2}e^{\sin 2}e^{-\sin t}. \nonumber \]
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
Integrated by Justin Marshall.