1. quadratic vs linear DifEq using LT

I was wondering if one was easier than the other or if it is situational.

Original problem was: y'' + 4y' + 5y = 39e^(t)sint and y(0) =-1 and y'(0) = -1

I took the LT resulting: $\displaystyle (s+5)(s-1) = 39/[(s-1)^2 +1] - s -5$. I was wondering about the denominator, since I could set it as
$\displaystyle s^2 - 2s +2$. And then I would have to multiply that by $\displaystyle - s - 5$. Would it be easier to do it as [(s-1)^2 +1]*[ - s -5 ]

2. Re: quadratic vs linear DifEq using LT

Originally Posted by SNAKE
I was wondering if one was easier than the other or if it is situational.

Original problem was: y'' + 4y' + 5y = 39e^(t)sint and y(0) =-1 and y'(0) = -1

I took the LT resulting: $\displaystyle (s+5)(s-1) = 39/[(s-1)^2 +1] - s -5$. I was wondering about the denominator, since I could set it as
$\displaystyle s^2 - 2s +2$. And then I would have to multiply that by $\displaystyle - s - 5$. Would it be easier to do it as [(s-1)^2 +1]*[ - s -5 ]
I can't edit for some reason, but there should be $\displaystyle L (s+5)(s-1) = 39/[(s-1)^2 +1] - s -5$

3. Re: quadratic vs linear DifEq using LT

Originally Posted by SNAKE
I was wondering if one was easier than the other or if it is situational.

Original problem was: y'' + 4y' + 5y = 39e^(t)sint and y(0) =-1 and y'(0) = -1

I took the LT resulting: $\displaystyle (s+5)(s-1) = 39/[(s-1)^2 +1] - s -5$. I was wondering about the denominator, since I could set it as
$\displaystyle s^2 - 2s +2$. And then I would have to multiply that by $\displaystyle - s - 5$. Would it be easier to do it as [(s-1)^2 +1]*[ - s -5 ]
I am sorry, it should say I just realized it is +5 so (s+5)(s-1) is incorrect.

4. Re: quadratic vs linear DifEq using LT

Originally Posted by SNAKE
I am sorry, it should say I just realized it is +5 so (s+5)(s-1) is incorrect.
do you still have a question about all this?

5. Re: quadratic vs linear DifEq using LT

yes, since it cannot be factored I guess it stays as it is on the left side. But on the right side I am still curious about the linear vs quadratic

6. Re: quadratic vs linear DifEq using LT

Originally Posted by SNAKE
yes, since it cannot be factored I guess it stays as it is on the left side. But on the right side I am still curious about the linear vs quadratic
You'll end up with the same answer either way. I think it's easier to just remember the ILT for a quadratic in s in the denominator.

Tables prefer a form of

$\dfrac {\omega_0} {(s+\alpha)^2+\omega_0^2}$

corresponding to

$e^{-at}\sin(\omega_0 t)u(t)$

there are a couple similar forms.

8. Re: quadratic vs linear DifEq using LT

For the inverse laplace, should I use e^at cos(bt)?

9. Re: quadratic vs linear DifEq using LT

Assuming we're going with what you've got at the bottom.

$L[y]=\dfrac{\alpha_1 s + \beta_1}{s^2+4s+5}+\dfrac{\alpha_2 s + \beta_2}{s^2-2s+2}$

I would rewrite it as

$\dfrac{\alpha_1 s + \beta_1}{(s+2)^2+1}+\dfrac{\alpha_2 s+\beta_2}{(s-1)^2+1}$

you can see with a bit of algebra these map to sinusoids times an exponential in the time domain as

$\dfrac {\omega_0}{(s+\gamma)^2+\omega_0^2}\rightarrow e^{-\gamma t}\sin(\omega_0 t)u(t)$

and

$\dfrac{s + \gamma}{(s+\gamma)^2+\omega_0^2}\rightarrow e^{-\gamma t}\cos(\omega_0 t)u(t)$

10. Re: quadratic vs linear DifEq using LT

If i factor out 2.79 I get 2.79*(.72s + 2)/ [[s+2]^2+1]. I don't see how I can use these. Can you please check my work? I solved the matrix by computer, but maybe somewhere I went wrong?

11. Re: quadratic vs linear DifEq using LT

Originally Posted by SNAKE
If i factor out 2.79 I get 2.79*(.72s + 2)/ [[s+2]^2+1]. I don't see how I can use these. Can you please check my work? I solved the matrix by computer, but maybe somewhere I went wrong?
For example the first term

$\dfrac {\alpha_1 s + \beta_1}{(s+2)^2+1} = \alpha_1\left(\dfrac {s + \frac {\beta_1}{\alpha_1}}{(s+2)^2+1}\right)=$

$\alpha_1\left(\dfrac {s + 2+\left(\frac {\beta_1}{\alpha_1}-2\right)}{(s+2)^2+1}\right)=$

$\alpha_1\left(\dfrac {s + 2}{(s+2)^2+1}\right)+\dfrac {\alpha_1}{\frac {\beta_1}{\alpha_1}-2}\left(\dfrac {1}{(s+2)^2+1}\right) \Rightarrow$

$\alpha_1 e^{-2t}\cos(t)u(t) + \dfrac {\alpha_1}{\frac {\beta_1}{\alpha_1} - 2} e^{-2t}\sin(t)u(t)$

see how it works?

Just do the exact same thing for the 2nd term.

12. Re: quadratic vs linear DifEq using LT

So you checked my work and I arrived where I did correctly? I see how you can manipulate the LT, but where do you get the u(t) from? On my table of LTs, there is no u(t).

13. Re: quadratic vs linear DifEq using LT

Originally Posted by SNAKE
So you checked my work and I arrived where I did correctly? I see how you can manipulate the LT, but where do you get the u(t) from? On my table of LTs, there is no u(t).
it shows in mine. It depends how you define the lower limit of the Laplace Transform as to whether you need the u(t).

I didn't check your work. You can go through it as well as I can, but if you just substitute your numbers into what I did below you should end up with the correct answers.

14. Re: quadratic vs linear DifEq using LT

If you wouldn't mind, could you please go over it. In my experience, the person who made the mistake is least likely to see it.

15. Re: quadratic vs linear DifEq using LT

Originally Posted by SNAKE
If you wouldn't mind, could you please go over it. In my experience, the person who made the mistake is least likely to see it.