Exact polynomial steps for a delay equation ($y^{\prime }(t)=ay(t-1),\varphi =1$)

Use the Malthusian delay differential equation with time measured in units of its constant delay:Here $a$ is a real constant. On the first interval $y(t-1)=1$, hence $y(t)=1+at$. On the next interval the delayed input is $1+a(t-1)$, already known, so the method of steps for a delay differential equation givesEach polynomial integrates the preceding interval's delayed polynomial, and its integration constant ensures continuity. At $a=1$, the step endpoints are $y(1)=2$, $y(2)=7/2$, and $y(3)=37/6$. At $t=3/2$, the value is $21/8$, but its derivative is the earlier value $y(1/2)=3/2$. This is the geometric distinction from the ordinary differential equation $y^{\prime }=y$, whose derivative would equal its current value.

More generally, for any integer $n\ge 0$,At an integer join the newly introduced term vanishes. Inside a step, differentiation lowers each power by one and gives exactly $a$ times the preceding step evaluated at $t-1$. Together with $y(0)=1$, this verifies the formula by mathematical induction; it is an exact construction, not a truncated power series.

The history and the new equation need not prescribe matching derivatives at the starting time; their mismatch can reappear later in higher derivatives.^[ref] In this example, $y^{\prime }(0^{-})=0$ and $y^{\prime }(0^{+})=a$. For $a\ne 0$, $y^{\prime }$ is continuous at $t=1$ but $y^{\prime \prime }$ jumps there by $a^2$; at $t=2$, the first jump is in $y^{\prime \prime \prime }$, of size $a^3$. The solution of a differential equation is therefore continuously differentiable for positive time, although it need not have continuous derivatives of every order at step boundaries.