6.4: General Solution of the Standard First Order Problem - an Alternative Derivation

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Consider again the general standard 1^st order problem, Equation 1.2.1, in which \(u(t)\) is the known input (excitation), \(x(t)\) is the output (response) that we seek, and \(a\) and \(b\) are constants, with \(a\) not necessarily negative:

\[\dot{x}-a x=b u(t), \text{ with IC } x\left(t_{1}\right) \text{ assumed known, find }x(t) \text{ for } t_{1} \leq t\label{eqn:6.15} \]

(We revert to the more general form now, rather than our stable standard 1^st order ODE, Equation 3.4.8 or Equation 6.2.1, in order to make this derivation more general, applicable for any physically realistic value or polarity of the constant \(a\).) This problem is a bit more general than we have considered previously, since we allow the initial time \(t_1\) to be different from zero. In order to expedite indexing of array quantities in computer algorithms, we use subscript 1, instead of 0, to denote the initial time and the initial value: \(x\left(t_{1}\right) \equiv x_{1}\).

To find a general solution of this 1^st order problem (ODE plus IC), we use the exponential function \(e^{a t}\) and closely related functions, for which we have the following basic identities:

\[e^{a t} \times e^{-a t_{1}}=e^{a\left(t-t_{1}\right)} \text { and } e^{a t} \times e^{-a t}=e^{0}=1\label{eqn:6.16} \]

We begin the general solution by multiplying the ODE by the integrating factor \(e^{-a t}\), recognizing that this will make the left-hand side a perfect derivative:

\[e^{-a t}(\dot{x}-a x=b u) \Rightarrow e^{-a t} \dot{x}-e^{-a t} a x=e^{-a t} b u \Rightarrow \frac{d}{d t}\left(e^{-a t} x\right)=e^{-a t} b u \nonumber \]

Now we integrate the multiplied equation from the initial time \(t_1\) to an arbitrary time instant \(t>t_{1}\), using \(\tau\) as the variable of integration:

\[\int_{\tau=t_{1}}^{\tau=t} \frac{d}{d \tau}\left[e^{-a \tau} x(\tau)\right] d \tau=\int_{\tau=t_{1}}^{\tau=t} e^{-a \tau} b u(\tau) d \tau \Rightarrow e^{-a t} x(t)-e^{-a t_{1}} x\left(t_{1}\right)=\int_{\tau=t_{1}}^{\tau=t} e^{-a \tau} b u(\tau) d \tau \nonumber \]

Finally, we multiply through by \(e^{a t}\), apply identities Equation \(\ref{eqn:6.16}\), move the IC term to the right-hand side, and arrive at the exact, general solution:

\[x(t)=e^{a\left(t-t_{1}\right)} x\left(t_{1}\right)+\int_{t=t_{1}}^{\tau=t} e^{a(t-\tau)} b u(\tau) d \tau, t_{1} \leq t\label{eqn:6.17} \]

The term \(e^{a t}\) need not be inside the integral, since it is not a function of the integration variable \(\tau\). Constant \(b\) also need not remain inside the integral. Solution Equation \(\ref{eqn:6.17}\) is comparable to, but more general than, the second form of convolution solution Equation 6.2.4.