Use the Laplace transform to solve the initial value problem.

Question

$$ y''-3y'+2y=e^{-t}; \quad\text{where}~ ~ y(2)=1, y'(2)=0 $$

Hint given: consider a translation of $y(x)$.

I am stuck on this problem on our homework. I don't understand what they mean by a "translation". Do they just mean a substitution?

Update - after spending a bit more time on this is where I'm stuck: $$ L(y)=\frac{\delta(s-1)}{s^2-3s+2} $$

Where do I go from here?

Answer 1

Firstly I can't see where you get the $\delta$-function from, please consider showing your working when you post a question here.

You want to use the following properties of the Laplace transform: $$\mathcal{L}[f'(t)]=sF(s)-f(0)$$ $$\mathcal{L}[f''(t)]=s^2F(s)-sf(0)-f'(0)$$ where $F(s)$ denotes the Laplace transform of $f(t)$. Unfortunately, you only have initial conditions for $t=2$. The idea is to translate $y$, i.e. define $w(t)=y(t+2)$. Can you see how this would be helpful?

Answer 2

1. You do not need to convert e^(-t) to a dirac-delta function. It is a dirac-delta function that is converted to e^(at) form to reduce the function.

2. Secondly, as we are supposed to work with initial conditions, which are not given here. we have,
   t=2 => consider n=t-2 so that n=0. => t=n+2
   now the given equation can be written as: y"(n+2)−3y'(n+2)+2y(n+2)=e^-(n+2)
   take u(n)=y(t)=y(n+2) => u'(n)=y'(n+2) and u"(n)=y"(n+2)
   also, u(0)=y(0+2)=1 and u'(0)=y'(0+2)=0

3. Hence the IVP under the given conditions becomes u"-3u'+2u=e^-(n+2)

you can now apply the laplace transform on both sides to get your answer

Answer 3

I cant understand where you get the $\delta$-function, too. According to my work: $$y''(t+2)-3y'(t+2)+2y(t+2)=e^{-(t+2)}\qquad y(0)=1\,,\,y'(0)=0$$ $$s^2\mathcal{L}[y(t+2)]-sy(0)-y'(0)-3(s\mathcal{L}[y(t+2)]-y(0))+2\mathcal{L}[y(t+2))]=e^{2s}\mathcal{L}[e^{-t}]$$ $$s^2(e^{2s}\mathcal{L}[y])-s-3(se^{2s}\mathcal{L}[y]-1)+2e^{2s}\mathcal{L}[y]=\frac{e^{2s}}{s+1}$$ $$e^{2s}(s^2-3s+2)\mathcal{L}[y]=\frac{e^{2s}}{s+1}+s-3$$ $$\mathcal{L}[y]=e^{-2s}(\frac{1}{(s-1)(s-2)})(\frac{e^{2s}}{s+1}+s-3)$$ $$\mathcal{L}[y]=e^{-2s}(\frac{1}{s-2}-\frac{1}{s-1})(\frac{e^{2s}}{s+1}+s-3)$$
The laplace inverse can be computed with convolution theorem. $$\mathcal{L}^{-1}[e^{-2s}(\frac{1}{s-2}-\frac{1}{s-1})]=e^{2(t-2)}-e^{(t-2)}$$ $$\mathcal{L}^{-1}[\frac{e^{2s}}{s+1}]=e^{-(t+2)}$$

I just don't know how to find the laplace inverse of $s$ and the calculate $s-3$ with the transform theorem. any help?

Answer 4

Let $f(t)=y(t+2)$. That way $f(0)=y(2)$ and $f'(0)=y'(2)$. Equivalently, $f(t-2)=y(t)$, $y'(t)=f'(t-2)$ and $y''(t)=f''(t-2)$. Then $$ f''(t-2)-3f'(t-2)+2f(t-2)=e^{-t},\;\;\; f(0)=1, f'(0)=0. $$ Substitute $t$ with $t+2$: $$ f''(t)-3f'(t)+2f(t) = e^{-t}e^{-2}, \;\;\; f(0)=1, f'(0)=0. $$