1. ## Delay differential equation

The problem is as follows:

$\frac{d}{dt}x(t) = x(t-1)$ with $x(t) = 1$ for $t \in [-1,0]$.

The solution is given by $x(t) = \sum_{k = 0}^{n} \frac{t-(k-1))^k}{k!}$ for $n-1 \leq t \leq n$.

Find the solution using the method of steps.

I have no idea what the method of steps is and I can't seem to find anything about it online. Any help/hints would be much appreciated!

2. ## Re: Delay differential equation

take a look at

http://www.mathfile.net/hicstat_fde.pdf

section 1.1.2

and yell back if that doesn't guide you to the solution

3. ## Re: Delay differential equation

Originally Posted by usersmuser
The problem is as follows:

$\frac{d}{dt}x(t) = x(t-1)$ with $x(t) = 1$ for $t \in [-1,0]$.

The solution is given by $x(t) = \sum_{k = 0}^{n} \frac{t-(k-1))^k}{k!}$ for $n-1 \leq t \leq n$.

Find the solution using the method of steps.

I have no idea what the method of steps is and I can't seem to find anything about it online. Any help/hints would be much appreciated!
I don't see why you would need to use a series solution when it can be solved exactly...

\displaystyle \begin{align*} \frac{\mathrm{d}x}{\mathrm{d}t} &= x\left( t - 1 \right) \\ \frac{1}{x} \,\frac{\mathrm{d}x}{\mathrm{d}t} &= t - 1 \\ \int{ \frac{1}{x}\,\frac{\mathrm{d}x}{\mathrm{d}t} \, \mathrm{d} t} &= \int{ \left( t - 1 \right) \,\mathrm{d}t} \\ \int{ \frac{1}{x}\,\mathrm{d}x} &= \frac{t^2}{2} - t + C_1 \\ \ln{ \left| x \right| } + C_2 &= \frac{t^2}{2} - t + C_1 \\ \ln{ \left| x \right| } &= \frac{t^2}{2} - t + C_1 - C_2 \\ \left| x \right| &= \mathrm{e}^{ \frac{t^2}{2} - t + C_1 - C_2 } \\ \left| x \right| &= \mathrm{e}^{C_1 - C_2} \,\mathrm{e}^{\frac{t^2}{2} - t} \\ x &= C \,\mathrm{e}^{\frac{t^2}{2} - t} \textrm{ where } C = \pm \mathrm{e}^{C_1 - C_2} \end{align*}

4. ## Re: Delay differential equation

You misunderstood the question.