Solve a very simple first order ODE in the form of an integral using Laplace Transforms. (5.4-5)

Question

I cannot duplicate the answer in the text book. Please examine my work to find the error.

Given:

Solve the following differential equation for $i(t)$ by Laplace Transforms. One initial condition $i(0)=0$ is given. The forcing function is only turned on when $t>1$.

$$100i+10\int i(t)dt=10(t-1)$$

My Setup:

Rewrite the equation dividing by 100 and apply the unit step function...

$$i+\frac{1}{10}\int i(t)dt=\frac{1}{10}(t-1)u(t-1)$$

My Solution Steps:

We now apply the Laplace transform across the entire equation giving us...

$$I+\frac{1}{10s}I=\frac{1}{10}\frac{e^{-s}}{s}$$

$$I(s+\frac{1}{10})=\frac{1}{10}e^{-s}$$

$$I=\frac{1}{10}e^{-s}\frac{1}{s+\frac{1}{10}}$$

Taking the inverse transform yields the following answer (for $t>1$)...

$$i(t)=\frac{1}{10}e^{-.1(t-1)}$$

Answer in Text:

$$i(t)=1-\frac{1}{10}e^{-.1(t-1)}$$

for $t>1$

My Question:

Where did I go wrong? I suspect my setup perhaps an error in applying the unit step to the forcing function? If that is the case can you show me why it went wrong and how to correct it? Thank you.

Answer 1

You didn't transform the right hand side correctly.

You have the (integral) equation $10 i(t) + \int_0^t i(\tau) d \tau = (t-1)u(t-1)$.

From the equation, we see that $\lim_{t \downarrow 1} i(t) = 0$, but the book answer gives ${ 9 \over 10}$. Hence the book answer cannot be correct.

Note that the right hand side is $t \mapsto t u(t)$, time shifted by 1, so the transform is $s \mapsto {1 \over s^2} e^{-s}$.

You have $10 \hat{i}(s) + {1 \over s} \hat{i}(s) = {1 \over s^2} e^{-s}$, and solving gives $\hat{i}(s) = {1 \over 10} {1 \over s}{1 \over s+ {1 \over 10}} e^{-s}$ from which we get $i(t) ={1 \over 10} \int_0^t e^{-{1 \over 10} (\tau-1)} u(\tau-1) d \tau$.

Computing this gives $i(t) = (1-e^{-{1 \over 10} (t-1)}) u(t-1)$.

Answer 2

I think you went wrong in your assertion that "applying the Laplace Transform across the whole equation [gives] us"

$$I + \frac{1}{10s}I = \frac{1}{10} \frac{e^{-s}}{s}$$

To simplify our problem, take the derivative of both sides wrt $t$ so we get an ODE and can use our Laplace Transform table. Hence we find

$$\begin{align} i'+\frac{1}{10} i &= \frac{d}{dt} \bigg(\frac{1}{10}(t - 1)u(t - 1) \bigg) \\\\ &= \frac{1}{10} u(t - 1) + \frac{1}{10} (t - 1) \delta (t - 1) \end{align}$$

Applying our transform, we find

$$\begin{align} sI(s) - i(0) + \frac{1}{10}I(s) &= I(s) \bigg(s + \frac{1}{10} \bigg) \\\\ &= \frac{1}{10} \frac{e^{-s}}{s} + \mathscr{L} \bigg[\frac{1}{10} t \delta (t-1) - \delta(t - 1) \bigg] \\\\ &= \frac{1}{10} \frac{e^{-s}}{s} + \frac{1}{10} e^{-s} - \frac{1}{10} e^{-s} \\\\ &= \frac{1}{10} \frac{e^{-s}}{s} \\\\ \implies I(s) &= \frac{1}{10} \frac{e^{-s}}{s \bigg(s + \frac{1}{10} \bigg)} \end{align}$$

Using partial fractions, we find

$$\begin{align} \frac{1}{s \bigg(s + \frac{1}{10} \bigg)} &= \frac{10}{s} - \frac{10}{s + \frac{1}{10}} \\ \implies I(s) &= \frac{e^{-s}}{s} - \frac{e^{-s}}{s + \frac{1}{10}} \end{align} $$

Transforming back, we find

$$I(t) = u(t - 1) - u(t - 1)e^{-\frac{1}{10} (t - 1)}$$

EDIT:

The problem was that you didn't transform your RHS correctly. The rule states that (see a Laplace Transform table)

$$\mathcal{L} [f(t - a)u(t - a)] = e^{-as}F(s)$$

where

$$F(s) = \mathcal{L} [f(t)]$$

In your case, we have

$$\frac{1}{10} (t - 1)u(t - 1) \\ \implies f(t) = t, a = 1$$

Now,

$$\begin{align} \mathcal{L} [t^{n}] &= \frac{n!}{s^{n + 1}} \\ \implies \mathcal{L} [f(t)] &= \mathcal{L} [t] \\ &= \frac{1}{s^{2}} \\ \implies \mathcal{L} [\frac{1}{10}(t - 1)u(t - 1)] &= \frac{1}{10} e^{-s} \cdot \frac{1}{s^{2}} \\ \end{align}$$