Lecture 3 - General Properties of First-Order Linear Differential Equations

Existence and uniqueness

If $p,g$ are continuous on an open interval $I$ containing $t_0$, then

$$y^{\prime }+p(t)y=g(t),y(t_0)=y_0$$

has exactly one solution on all of $I$.

Existence. Set $P(t)={\int }_{t_0}^{t}p(s)ds$. The integrating-factor formula gives

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

Continuity of $p,g$ makes this expression differentiable. The product rule and fundamental theorem of calculus give $y^{\prime }=-py+g$, and evaluation at $t_0$ gives $y(t_0)=y_0$.

Uniqueness. If $y_1,y_2$ solve the same IVP, their difference $w=y_1-y_2$ satisfies

$$w^{\prime }+pw=0,w(t_0)=0.$$

Therefore

$$w(t)=w(t_0)e^{-P(t)}=0,$$

so $y_1=y_2$ throughout $I$.

More generally, solutions with different initial values satisfy

$$y_1(t)-y_2(t)=[y_1(t_0)-y_2(t_0)]e^{-P(t)}.$$

Since $e^{-P(t)}>0$, distinct solution curves cannot intersect on $I$.

Singular coefficients and solution intervals

Continuity of the coefficients is a sufficient condition for the theorem. If it fails, examine the equation and its domain directly.

For

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

the solution is $y=-3t/2$ on $(-\infty ,0)$. Its formula extends smoothly through zero, but the differential equation is undefined there.

For

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

the solution is $y=-6/t$ on $(-\infty ,0)$, and diverges as $t\to 0^{-}$.

A formula extending past a singularity does not make it a solution at a point where the equation is undefined.

Superposition

Define the linear operator

$$L[y]=y^{\prime }+p(t)y.$$

For constants $c_1,c_2$,

$$L[c_1{y}_1+c_2{y}_2]=c_{1}L[y_1]+c_{2}L[y_2].$$

If $L[y_1]=L[y_2]=0$, then every linear combination $c_1{y}_1+c_2{y}_2$ is another homogeneous solution.

If $L[y_p]=g$ and $L[y_h]=0$, then

$$L[y_p+C{y}_h]=g.$$

Conversely, any solution $y$ of $L[y]=g$ differs from $y_p$ by a homogeneous solution. Choosing a nonzero homogeneous solution $y_h$ therefore gives the general solution

$$\boxed{y=y_p+C{y}_h.}$$

The homogeneous solution space is one-dimensional; the nonhomogeneous solution set is a translate of that space.

For two solutions of the same nonhomogeneous equation,

$$L[y_1+y_2]=2g,L[y_1-y_2]=0.$$

Their sum solves the original equation only if $g\equiv 0$. Their difference always solves the homogeneous equation.

Example

$$y^{\prime }=y-t+1.$$

A particular solution is $y_p=t$, since $1=t-t+1$. The homogeneous equation $y_h^{\prime }=y_h$ has the nonzero solution $e^t$, so

$$\boxed{y=t+C{e}^t.}$$

Why linearity matters

For the nonlinear equation

$$y^{\prime }=y^2-t^2+1,$$

$y_p=t$ is a solution, and $z=-1/t$ solves $z^{\prime }=z^2$ on intervals avoiding zero. Their combination $y=t-c/t$ gives

$$\begin{array}{rl}{y}^{\prime }& =1+\frac{c}{t^2},\\ y^2-t^2+1& =1-2c+\frac{c^2}{t^2}.\end{array}$$

Equality on an interval would require

$$2c{t}^2+c-c^2=0$$

for every $t$, which forces $c=0$. Superposition does not hold in general for nonlinear equations.

Previous: Lecture 2 - Linear First-Order Differential Equations