Lecture 2 - Linear First-Order Differential Equations

Linear first-order equations

A first-order equation is linear if it can be written

$$\boxed{y^{\prime }+p(t)y=g(t).}$$

It is homogeneous when $g\equiv 0$ and nonhomogeneous otherwise.

Equation	Classification
$y^{\prime }=t{y}^2$	Nonlinear
$y^{\prime }=t^{2}y$	Linear, homogeneous
$(y^{\prime }{)}^2=ty$	Nonlinear
$(\cos t)y^{\prime }+e^{t}y=\sin t$	Linear, nonhomogeneous

The last equation has standard form

$$y^{\prime }+\frac{e^t}{\cos t}y=\tan t$$

on intervals where $\cos t\ne 0$.

Homogeneous solution

Let $P^{\prime }(t)=p(t)$. Since

$${\left(e^{-P(t)}\right)}^{\prime }=-p(t)e^{-P(t)},$$

the homogeneous equation has solutions

$$\boxed{y_h(t)=C{e}^{-P(t)}.}$$

These are all the solutions: multiplying $y^{\prime }+py=0$ by $e^P$ gives $(e^{P}y{)}^{\prime }=0$, hence $e^{P}y=C$.

Integrating factor

For continuous $p,g$ on an interval, multiply by the integrating factor $\mu =e^P$:

$$\begin{array}{rl}{y}^{\prime }+py& =g,\\ e^P{y}^{\prime }+p{e}^{P}y& =e^{P}g,\\ (e^{P}y{)}^{\prime }& =e^{P}g.\end{array}$$

Thus

$$\boxed{y(t)=e^{-P(t)}\left(C+\int e^{P(t)}g(t)dt\right),}$$

where the integral denotes one antiderivative and $C$ supplies the arbitrary constant.

For $y(t_0)=y_0$, choose

$$P(t)={\int }_{t_0}^{t}p(s)ds.$$

Then $P(t_0)=0$, and integrating $(e^{P}y{)}^{\prime }$ from $t_0$ to $t$ gives

$$\boxed{y(t)=e^{-P(t)}\left[y_0+{\int }_{t_0}^t{e}^{P(s)}g(s)ds\right].}$$

The outer factor depends on $t$; the integrand depends on $s$. They cannot be cancelled.

Example

$$y^{\prime }-\frac{2t}{1+t^2}y=2t,y(1)=3.$$

The general solution, using $P=-\ln (1+t^2)$, is

$$y=(1+t^2)\left[C+\ln (1+t^2)\right].$$

For the IVP, normalize at $t_0=1$:

$$\begin{array}{rl}P(t)& ={\int }_1^t\frac{-2s}{1+s^2}ds=\ln 2-\ln (1+t^2),\\ e^{P(t)}& =\frac{2}{1+t^2},\\ {\int }_1^t{e}^{P(s)}g(s)ds& ={\int }_1^t\frac{4s}{1+s^2}ds=2\ln \frac{1+t^2}{2}.\end{array}$$

Therefore

$$\boxed{y(t)=\frac{1+t^2}{2}\left[3+2\ln \frac{1+t^2}{2}\right].}$$

At $t=1$, the logarithm vanishes and $y(1)=3$.

Variation of parameter

Start with the nonzero homogeneous solution $y_h=e^{-P(t)}$ and allow its coefficient to vary:

$$y=C(t)y_h(t).$$

Substitution gives

$$\begin{array}{rl}{y}^{\prime }+py& =C^{\prime }{y}_h+C(y_h^{\prime }+p{y}_h)\\ & =C^{\prime }{y}_h=g.\end{array}$$

Hence

$$C^{\prime }=\frac{g}{y_h}=e^{P}g,y=e^{-P}\left(C_0+\int e^{P}gdt\right).$$

This recovers the integrating-factor formula. The same idea extends to linear systems.

Special cases and long-term behavior

For the homogeneous IVP,

$$y(t)=y_0\exp \left(-{\int }_{t_0}^{t}p(s)ds\right).$$

Zero initial data gives $y\equiv 0$. Nonzero solutions keep their sign, and

$$\frac{d}{dt}\ln |y|=-p(t).$$

Thus $p>0$ decreases the magnitude and $p<0$ increases it. For $y_0\ne 0$, the limits depend on the accumulated integral:

$$\begin{array}{rl}{\int }_{t_0}^{t}p(s)ds\to +\infty & ⟹y(t)\to 0,\\ {\int }_{t_0}^{t}p(s)ds\to -\infty & ⟹|y(t)|\to \infty .\end{array}$$

The sign alone does not determine the limit. For example, $p(t)=e^{-t}$ with $t_0=0$ gives

$$y(t)=y_0{e}^{e^{-t}-1}⟶y_0/e.$$

Constant coefficient

$$y^{\prime }=ay,y(t_0)=y_0⟹\boxed{y(t)=y_0{e}^{a(t-t_0)}.}$$

Power law

On $t>0$,

$$y^{\prime }-\frac{a}{t}y=0,P=-a\ln t⟹\boxed{y=C{t}^a.}$$

Constant forcing

For $y^{\prime }=ay+b$ with $a\ne 0$, the equilibrium is $y_{\ast }=-b/a$. Set $z=y-y_{\ast }$:

$$z^{\prime }=az⟹\boxed{y(t)=y_{\ast }+(y_0-y_{\ast })e^{a(t-t_0)}.}$$

If $a<0$, every solution converges to $y_{\ast }$. If $a>0$, every non-equilibrium solution moves away from it. If $a=0$, $y=y_0+b(t-t_0)$.

Piecewise coefficients

If $p$ or $g$ changes formula at $t=c$, solve the linear IVP on each interval and match the value of $y$ at the switch:

$$y_c=\underset{t\to c^{-}}{\lim }y(t),y(c)=y_c.$$

Use this as the initial condition for the second interval. The solution is continuous; its derivative may jump. At a derivative jump, the DE holds on each open piece rather than as a classical equation at the switch.

For example, if $y^{\prime }+y=g(t)$, $y(0)=y_0$, and $g=g_1$ before $c>0$, $g=g_2$ afterward, with $g_1,g_2$ constant, then

$$y(t)=\{\begin{array}{ll}{g}_1+(y_0-g_1)e^{-t},& 0\le t<c,\\ g_2+(y_c-g_2)e^{-(t-c)},& t\ge c,\end{array}{y}_c=g_1+(y_0-g_1)e^{-c}.$$

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