# 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

## 1 Answer

The constants would be the initial conditions, such as an initially charged capacitor or an inductor with an initial current. If all constants are set to zero it means that you have zero initial conditions, which is usually the standard assumption unless stated otherwise.

• Thank you. You mean that I have to add any constants when the system has initial conditions? – Danny_Kim Nov 6 '15 at 10:50
• @Danny_Kim: Yes, that's right. – Matt L. Nov 6 '15 at 12:45