differential equation solved by Laplace transform

Submitted by Structure on Sun, 10/19/2008 - 21:19.

We solve the differential equation

$y"(t)+y(t)=f(t)$

with initial condition

$y(0)=0,\:y'(0)=0$

where

$f(t)=u(t-2\pi)-u(t-\pi)$

$\mathcal{L}(y(t)(s)=Y(s)$

$\mathcal{L}(y'(t)(s)=sY(s)$

$\mathcal{L}(y"\;(t)(s)=s^2Y(s)$

We have

$\mathcal{L}(y"(t)+y(t))(s)=s^2Y(s)+Y(s)=F(s)=\frac{1}{s}(e^{-2\pi s}-e^{-\pi s})$

We have

$Y(s)=\frac{1}{s^2+1}F(s)$

We have from convolution theorem

$y(t)=\sin*f(t)=\int_{0}^{t}f(\tau)\sin(t-\tau)d\tau$

proof without words

Submitted by Structure on Sun, 10/05/2008 - 17:55.

Submitted by Structure on Wed, 09/24/2008 - 06:18.

We solve the equation

$a\sin x+b\cos x+c=0$

where $a,b,c \in R$ and $a^2+b^2>0$ .
Then we can write

$\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x+\frac{c}{\sqrt{a^2+b^2}})=0$

There is a unique $t\in[0,2\pi)$ such as

$\cos t=\frac{a}{\sqrt{a^2+b^2}}$

and

$\sin t=\frac{b}{\sqrt{a^2+b^2}}$

.
so we can write

$\sqrt{a^2+b^2}(\sin x\cos t+\cos x\sin t+\frac{c}{\sqrt{a^2+b^2}})=0$

$\sqrt{a^2+b^2}(\sin (x+t)+\frac{c}{\sqrt{a^2+b^2}})=0$

Now, if

$|\frac{c}{\sqrt{a^2+b^2}}|\leq 1$

we have a countable set of solutions