Using Laplace transforms to solve differential equations - with a twist

Laura W
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I've been given this:
x''+ x = 4δ(t-2π)

The question asks:
With initial conditions of x(0) = 1 and x'(0) = 0, find x(t) using Laplace transforms.

I can easily get it to this:

4(sin(t-2π)u(t-2π))

But the question says "express your final solution without use of the unit step function". This is where I get confused as I'm not quite sure as to how to do that. Will it just be 4sin(t)? Considering the sin function repeats at 2π.
 
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Laura W said:
I've been given this:
x''+ x = 4δ(t-2π)

The question asks:
With initial conditions of x(0) = 1 and x'(0) = 0, find x(t) using Laplace transforms.

I can easily get it to this:

4(sin(t-2π)u(t-2π))

But the question says "express your final solution without use of the unit step function". This is where I get confused as I'm not quite sure as to how to do that. Will it just be 4sin(t)? Considering the sin function repeats at 2π.

In general if you have ##x=f(t)u(t-a)## you would write a two piece function since ##u(t-a) = 0## if ##t<a## and ##u(t-a) = 1## if ##t>a##. So you would say$$
x =\left\{\begin{array}{l}
0,~t<a\\
f(t),~t>a
\end{array}\right.$$Does that help?
 
LCKurtz said:
In general if you have ##x=f(t)u(t-a)## you would write a two piece function since ##u(t-a) = 0## if ##t<a## and ##u(t-a) = 1## if ##t>a##. So you would say$$
x =\left\{\begin{array}{l}
0,~t<a\\
f(t),~t>a
\end{array}\right.$$Does that help?

That's what I was thinking but I was wondering if there was another way of transforming it at all? Seems to simple of an answer for this specific lecturer who wrote the question.
 
Laura W said:
I've been given this:
x''+ x = 4δ(t-2π)

The question asks:
With initial conditions of x(0) = 1 and x'(0) = 0, find x(t) using Laplace transforms.

I can easily get it to this:

4(sin(t-2π)u(t-2π))

But the question says "express your final solution without use of the unit step function". This is where I get confused as I'm not quite sure as to how to do that. Will it just be 4sin(t)? Considering the sin function repeats at 2π.
Your answer isn't correct. Did you forget to incorporate the initial conditions?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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