= Linear differential equation
{title2=$L[y]=g(t)$}
{wiki}
= Linear ODE
{synonym}
A scalar <linear differential equation> of order $n\geq1$ has the form
$$
L[y](t)=a_n(t)y^{(n)}(t)+a_{n-1}(t)y^{(n-1)}(t)+\cdots+a_1(t)y'(t)+a_0(t)y(t)=g(t).
$$
Here the <coefficients> $a_j$ and the <forcing term> $g$ are prescribed <functions> of the independent variable $t$. A classical <solution of a differential equation> is an $n$-times <differentiable function> satisfying the identity at every point of its <interval>. The unknown <function> $y$ and its <derivatives> enter to the first power, without products between them or <coefficients> depending on $y$. Equivalently, the <differential operator> $L$ is a <linear map>: for constant <scalars> $\alpha,\beta$,
$$
L[\alpha u+\beta v]=\alpha L[u]+\beta L[v].
$$
The <forcing term> may be zero, giving a <homogeneous linear differential equation>, or nonzero, giving an <inhomogeneous linear differential equation>. <Linearity> of $L$ does not mean that every <function graph> of a <solution of a differential equation> is a straight line.
To use the usual order-$n$ <initial value problem> theory on an <interval> $I$, assume that $a_n(t)\ne0$ throughout $I$ and that all $a_j$ and $g$ are <continuous functions> there. Dividing by $a_n$ normalizes the highest <derivative>; specifying $y(t_0),y'(t_0),\ldots,y^{(n-1)}(t_0)$ at $t_0\in I$ then selects one <solution of a differential equation> throughout $I$. A zero of $a_n$ needs separate analysis: one cannot divide through it and retain these conclusions. The first-order normalization appears in https://openstax.org/books/calculus-volume-2/pages/4-5-first-order-linear-equations[OpenStax's definition and standard form of linear equations]; the corresponding system theorem is in https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/8ae57eb392d85ce692965b97dda608fc_MIT18_03SCF11_s35_4text.pdf[MIT's existence and uniqueness notes for linear systems].
The <coefficients> need not be <linear functions> of $t$. For example, $t^2y'+y=0$ is a <linear differential equation> on $t>0$, although $t^2$ is quadratic. Its <solution of a differential equation> with <initial condition> $y(1)=1$ is $y(t)=e^{1/t-1}$: its <derivative> is $y'=-y/t^2$, so substitution gives $t^2y'+y=0$, and $y(2)=e^{-1/2}\approx0.60653$. The zero of the highest <coefficient> at $t=0$ explains the domain restriction. In contrast, $y'+y^2=0$, $yy'+y=0$ and $y'+\sin y=0$ are <nonlinear ordinary differential equations> because of their dependence on the unknown <function>. Variable <coefficients> and <nonlinear ordinary differential equations> are different issues; see https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/acc65618b35025a2e9db939ec35b1de5_MIT18_03SCF11_s4_1text.pdf[MIT's definitions and examples of first-order linear equations].
The practical benefit of <linearity> is that it separates the effect of the <initial condition> from the effect of the <forcing term>. Once one <particular solution> $y_p$ satisfies $L[y_p]=g$, every <solution of a differential equation> is
$$
y=y_p+h,\qquad h\in\ker L.
$$
Here $\ker L$ is the <kernel of a linear map>, consisting of <solutions of a differential equation> to the associated <homogeneous linear differential equation>. Indeed, $L[y-y_p]=g-g=0$, and conversely $L[y_p+h]=g+0=g$. Thus the family for a fixed <forcing term> is an <affine space> obtained by translating $\ker L$. If $g$ is not identically zero, the zero <function> is absent, so this family is not a <vector space> under the usual operations on <functions>. This short argument explains why a <general solution> is a <particular solution> plus a <homogeneous solution>.
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