Solution to $y′(t) + y(t − 1) = \cos^{2}\pi t$

Question

Determine a solution with period 2 of the differential-difference equation $y′(t) + y(t − 1) = \cos^{2}\pi t$.

Solution:

period: $T = 2 \Rightarrow \Omega = \frac{2 \pi}{2} = \pi$

Fourier series: $y(t) \sim \sum _{n=1}^{\infty} y_ne^{i \pi nt}$

Can anyone point me in the right direction from this? I have no working examples to study or anything so I am kinda stuck.

Answer 1

First, we find the solution of the non-difference equation:

$$y'(t) + y(t) = \cos^2(\pi t) = \frac 12 \left( 1 + \cos 2\pi t \right)$$

• Homogeneous solution $y_h(t) = Ce^{-t}$
• Solution of the inhomogeneous equation $y' + y = 1/2$: $\displaystyle y_1(t) = \frac 12$
• Solution of the inhomogeneous equation $y' + y = \frac 12 \cos 2\pi t$: $\displaystyle y_2(t) = \frac{1}{2 + 8\pi^2}\left( 2\pi\sin(2\pi t) + \cos(2\pi t) \right) $

Hence the general solution of $y' + y = \cos^2(\pi t)$ is

$$y(t) = Ce^{-t} + \frac 12 + \frac{1}{2 + 8\pi^2}\left( 2\pi\sin(2\pi t) + \cos(2\pi t) \right)$$

Now how can we modify the equation so that it satisfies the original equation. Start with $C = 0$.

Then $$y'(t) + y(t-1) = y'(t) + y(t)$$ as $\sin(2\pi t)$ and $\cos(2\pi t)$ have period $1$ and hence

$$y(t) = \frac{1}{2 + 8\pi^2}\left( 2\pi\sin(2\pi t) + \cos(2\pi t) \right) + \frac 12$$

is a solution which has period $1$ and period $2$. (Or period $n$ for any $n \in\mathbb N$.)