# Why ignores constant when applying integral into differential-equation.?

I studied how to draw 'Direct-form I' and 'Direct-form II' from differential equation.

The book taught me like this:

$$\mbox{Give equation: }{a_0}{y(t)}+{a_1}{y'(t)}+{a_2}{y''(t)} = {b_0}{x(t)}+{b_1}{x'(t)}+{b_2}{x''(t)}$$

step 1. change from differentiator into integrator, because integrator is cheaper than differentiator.

\begin{array}{} {a_0}{y^{(2)}(t)}+{a_1}{y^{(1)}(t)}+{a_2}{y(t)} = {b_0}{x^{(2)}(t)}+{b_1}{x^{(1)}(t)}+{b_2}{x(t)} & \begin{cases} x^{(i)} \mbox{ is i-times integral of x}\\ y^{(i)} \mbox{ is i-times integral of y} \end{cases} \end{array}

step 2. change form like this: $y(t)=...$

\begin{array}{} \displaystyle y(t)=\frac{1}{a_2}\left( {b_0}{x^{(2)}(t)}+{b_1}{x^{(1)}(t)}+{b_2}{x(t)}-{a_0}{y^{(2)}(t)}-{a_1}{y^{(1)}(t)}\right) \end{array}

My question is why doesn't consider constant value in step 1?

• Are you sure that you are not confusing differentiation and delay? The standard use for a "Direct Form I" would be a difference equation and the continuous equivalent might be closer to $a_0\cdot y(t) + a_1 \cdot y(t-T) + a_2 \cdot y(t - 2T) = ...$ – Hilmar Nov 6 '15 at 19:48
• It is not confused. Discrete time is easier for me to understand than continuous time. – Danny_Kim Nov 6 '15 at 20:03