MHB Yes, that is the correct answer. Good job!

[shamieh]

When the Laplace Transform is applied to a certain IVP, the resulting equation is $(s^2Y - 2s - 1) + (sY - 2) - 2Y = \frac{4}{s}$

What was the IVP?

So I think I've solved this, but just want to make sure I got the correct answer.

I got:

$y(0) = 2$ , $y'(0) = 1$, & $C = 2$

$\therefore y'' + y' - 2y = 4$ was the IVP correct?

 

[Ackbach]

When the Laplace Transform is applied to a certain IVP, the resulting equation is $(s^2Y - 2s - 1) + (sY - 2) = \frac{4}{s}$

What was the IVP?

So I think I've solved this, but just want to make sure I got the correct answer.

I got:

$y(0) = 2$ , $y'(0) = 1$, & $C = 2$

$\therefore y'' + y' + 2y = 4$ was the IVP correct?

Hmm. We have that
\begin{align*}
L\{y(t)\}&=Y(s) \\
L\{y'(t)\}&=s Y(s)-y(0) \\
L\{y''(t)\}&=s^2 Y(s) -sy(0)-y'(0).
\end{align*}
Now, our ansatz will be
$$(s^2Y - 2s - 1) + (sY - 2)=L\{A y''(t)+By'(t)+Cy(t)\}.$$
The absence of any $Y$ in the LT suggests that $C=0$. Taking the LT on the RHS, then, yields
$$A(s^2 Y-sy(0)-y'(0))+B(s Y-y(0)).$$
It must be that $A=1$, or the $s^2Y$ term wouldn't work. That is, our equation is now
$$(s^2Y - 2s - 1) + (sY - 2)=s^2 Y-sy(0)-y'(0)+B(s Y-y(0)).$$
Is there a value of $B$ that would make this work? Well, let's see if we can determine $y(0)$ and $y'(0)$ first. On the LHS, there is the $-2s$ term - this must mean that $-y(0)=-2$, or $y(0)=2$. Plugging this into the equation yields
$$ - 1 + sY - 2=-y'(0)+B(s Y-2).$$
The only value of $B$ that works here is $B=1$. Thus, our equation boils down to
$$ -3=-y'(0)-2.$$
This implies $y'(0)=1$. So the IVP must be
\begin{align*}
y''(t)+y'(t)&=4 \\
y(0)&=2 \\
y'(0)&=1.
\end{align*}
We double-check by taking the LT:
\begin{align*}
s^2 Y -sy(0)-y'(0)+s Y-y(0)&=\frac4s \\
s^2 Y-2s-1+sY-2&=\frac4s.
\end{align*}
Can you see where you went wrong?

 

[shamieh]

Ach, I wrote the incorrect question. It should have said $(s^2Y - 2s - 1) + (sY - 2) -2Y = \frac{4}{s}$ instead of $(s^2Y - 2s - 1) + (sY - 2) = \frac{4}{s}$ would my answer then be correct?

 

[shamieh]

I've edited the question to what I meant to have it say in the first place I'm still unsure if my answer is correct tho. Can you verify?

 

[Ackbach]

Yes, I should say your answer is now correct. The $-2Y$ term can only come from a $-2y(t)$ term in the original DE. The rest of the analysis goes through as is.