laplace differential equation with conditions

Question

I have to solve this differential equation with laplace
$y'' + 6y' + 9y = \begin{cases}5t & 0 < t \le 3 \\ 0 & t>= 3\end{cases}$ and $y(0)=1, y'(0)=1$

I know what to do with the left part of the equation but with the right i don't know.

Could you help me with that.

Thank you
EDIT
I found this result for Y(S) :

For the first term by using translation theorem i found this answer :

wich gives :

But the answer from my TI nspire is :

Do you know why i have an extra u(t-3)? thank you

Answer 1

Hint

$$y''+6y'+9y=5t-5t\,\,{{u}_{3}}(t)$$ $$s^2F(s)-sf(0)-f'(0)+6F(s)-6f(0)+9F(s)=\frac{5}{s^2}-\frac{5e^{-3s}}{s^2}-\frac{15e^{-3s}}{s}$$

Answer 2

First you must know how to rewrite a piecewise defined function in terms of unit step functions. I will illustrate with a function defined in three pieces.

Let $f(x)=\begin{cases}y_0(t)\text{ for }0<t\le a\\ y_1(t)\text{ for }a<t\le b\\ y_2(t)\text{ for }t\ge b \end{cases}$

Then $f(t)=y_0(t)\,u_0(t)+[y_1(t)-y_0(t)]\,u_a(t)+[y_2(t)-y_1(t)]\,u_b(t)$

Following this general pattern

If $f(t)=\begin{cases}5t & \text{ for }0 < t \le 3 \\ 0 &\text{ for } t\ge 3\end{cases}$ then $f(t)=5t\,u_0(t)+(0-5t)\,u_3(t)=5t-5t\,u_3(t)$ since $u_0(t)=1$ for $t>0$.

You must also be aware of the following Laplace transform:

\begin{equation} \mathcal{L}\{g(t)\,u_a(t)\}=\mathcal{L}\{g(t+a)\}e^{-as} \end{equation}

Therefore $\mathcal{L}\{5t\,u_3(t)\}=\mathcal{L}\{5(t+3)\}e^{-3s}=\dfrac{5}{s^2}e^{-3s}+\dfrac{15}{s}e^{-3s}$

I assume you can do the rest.