Second order differential help (Laplace)

[Bibbster]

Hey guys , having a bit of bother getting a solution for this question. Any help would be greatly appreciated!
There is a Picture attached showing how far I have got ..View attachment 1586

 

[Dustinsfl]

Re: Second order differential help (laplace)

How I would do is solve the homogeneous which is relatively trivial and you get
\[
y_c = e^{-\frac{1}{2}t}\left[c_1\cos\left(t\frac{\sqrt{15}} {2}\right) + c_2\sin\left(t\frac{\sqrt{15}} {2}\right)\right]
\]
Then let \(y_p = at^2 + bt + c + d\cos(2t) + e\sin(2t)\).
Then take \(y_p'' + y_p' + 4y_p = \text{RHS}\), equate the coefficients, and then use linear alg to find the coefficients of the paticular solution.

---

I just saw you wanted a Laplace transform.

When I apply \(\mathcal{L}\), I get
\[
s^2y(s) - s - 2 + sy(s) - 1 + 4y(s) = \frac{8}{s^3} + \frac{2}{s^2} + \frac{2}{s} + \frac{s}{s^2 + 4} + \frac{2}{s^2 + 4}
\]
which simplifies to
\[
(s^2 + s + 4)y(s) - s - 3 = \text{RHS}.
\]
I don't think we have the same here.

 

[Chris L T521]

Re: Second order differential help (laplace)

How I would do is solve the homogeneous which is relatively trivial and you get
\[
y_c = e^{-\frac{1}{2}t}(c_1\cos(t\sqrt{15}) + c_2\sin(t\sqrt{15}))
\]
Then let \(y_p = at^2 + bt + c + d\cos(2t) + e\cos(2t)\).
Then take \(y_p'' + y_p' + 4y_p = \text{RHS}\), equate the coefficients, and then use linear alg to find the coefficients of the paticular solution.

Just letting you know that you made a minor mistake in your $y_p$; you accidentally put $\cos(2t)$ twice (i.e. $d\cos(2t)+e\cos(2t)$ when I'm sure you meant $d\cos(2t)+e\sin(2t)$.

 

[Bibbster]

Re: Second order differential help (laplace)

How I would do is solve the homogeneous which is relatively trivial and you get
\[
y_c = e^{-\frac{1}{2}t}\left[c_1\cos\left(t\frac{\sqrt{15}} {2}\right) + c_2\sin\left(t\frac{\sqrt{15}} {2}\right)\right]
\]
Then let \(y_p = at^2 + bt + c + d\cos(2t) + e\sin(2t)\).
Then take \(y_p'' + y_p' + 4y_p = \text{RHS}\), equate the coefficients, and then use linear alg to find the coefficients of the paticular solution.

---

I just saw you wanted a Laplace transform.

When I apply \(\mathcal{L}\), I get
\[
s^2y(s) - s - 2 + sy(s) - 1 + 4y(s) = \frac{8}{s^3} + \frac{2}{s^2} + \frac{2}{s} + \frac{s}{s^2 + 4} + \frac{2}{s^2 + 4}
\]
which simplifies to
\[
(s^2 + s + 4)y(s) - s - 3 = \text{RHS}.
\]
I don't think we have the same here.

I see where the confusion is coming from , my mistake v(o)=-1 rather than 1 as it says in above picture. Sorry about that! However, was I along the right lines in my above working or?

Thanks :)

 

[Bibbster]

Re: Second order differential help (laplace)

How I would do is solve the homogeneous which is relatively trivial and you get
\[
y_c = e^{-\frac{1}{2}t}\left[c_1\cos\left(t\frac{\sqrt{15}} {2}\right) + c_2\sin\left(t\frac{\sqrt{15}} {2}\right)\right]
\]
Then let \(y_p = at^2 + bt + c + d\cos(2t) + e\sin(2t)\).
Then take \(y_p'' + y_p' + 4y_p = \text{RHS}\), equate the coefficients, and then use linear alg to find the coefficients of the paticular solution.

---

I just saw you wanted a Laplace transform.

When I apply \(\mathcal{L}\), I get
\[
s^2y(s) - s - 2 + sy(s) - 1 + 4y(s) = \frac{8}{s^3} + \frac{2}{s^2} + \frac{2}{s} + \frac{s}{s^2 + 4} + \frac{2}{s^2 + 4}
\]
which simplifies to
\[
(s^2 + s + 4)y(s) - s - 3 = \text{RHS}.
\]
I don't think we have the same here.

I apologize for being a nuisance, had a look at this when i got home from work again and noticed the following:

$$\displaystyle \dfrac{2s}{(s^2+4)(s^2+s+4)} \ = \ \dfrac{2s}{(s^2+4)} + \dfrac{2s}{(s^2+s+4)}$$

is incorrect ^^^ as i had in initial workings.

Following on from this i had an idea that if i split up the first term i could get: $$\displaystyle \ \dfrac{-s}{(s^2+s+4)} + \dfrac{1}{(s^2+s+4)}$$ which would then leave me to find possible solutions for $$\displaystyle (s^2+s+4)$$ with one being $$\displaystyle =(s+0.5)^2+15/4$$. Then after that (if correct) I'm once again confused . :(

below are my new workings to keep you guys on track with what I am thinking. (first term isn't split up as unsure ).

View attachment 1593

 

[Dustinsfl]

Re: Second order differential help (laplace)

I apologize for being a nuisance, had a look at this when i got home from work again and noticed the following:

$$\displaystyle \dfrac{2s}{(s^2+4)(s^2+s+4)} \ = \ \dfrac{2s}{(s^2+4)} + \dfrac{2s}{(s^2+s+4)}$$

is incorrect ^^^ as i had in initial workings.

Following on from this i had an idea that if i split up the first term i could get: $$\displaystyle \ \dfrac{-s}{(s^2+s+4)} + \dfrac{1}{(s^2+s+4)}$$ which would then leave me to find possible solutions for $$\displaystyle (s^2+s+4)$$ with one being $$\displaystyle =(s+0.5)^2+15/4$$. Then after that (if correct) I'm once again confused . :(

below are my new workings to keep you guys on track with what I am thinking. (first term isn't split up as unsure ).

View attachment 1593

That fraction you are speaking of reduces to
\[
\frac{2}{s^2+4}-\frac{2}{s^2+s+4}
\]

 

[Bibbster]

Re: Second order differential help (laplace)

That fraction you are speaking of reduces to
\[
\frac{2}{s^2+4}-\frac{2}{s^2+s+4}
\]

as you said , i now have $$\dfrac{2}{(s^2+4)}$$$$-$$$$\dfrac{2}{(s^2+s+4)}$$ with $$\dfrac{2}{(s^2+4)}$$ becoming $$Sin2t$$ and $$\dfrac{2}{(s^2+s+4)}$$ becoming $$\dfrac{2}{(s^2+0.5)^2+15/4}$$ however I am struggling to tackle this one , if you could point me in right direction that'd be great

 

[Chris L T521]

Re: Second order differential help (laplace)

as you said , i now have $$\dfrac{2}{(s^2+4)}$$$$-$$$$\dfrac{2}{(s^2+s+4)}$$ with $$\dfrac{2}{(s^2+4)}$$ becoming $$Sin2t$$ and $$\dfrac{2}{(s^2+s+4)}$$ becoming $$\dfrac{2}{(s^2+0.5)^2+15/4}$$ however I am struggling to tackle this one , if you could point me in right direction that'd be great

Now recall that $\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t)$

Hence,

\[\mathcal{L}^{-1}\left\{\frac{2}{\left(s +\frac{1}{2}\right)^2 +\frac{15}{4}}\right\}= e^{-\frac{1}{2}t}\mathcal{L}^{-1}\left\{\frac{2}{s^2+\frac{15}{4}}\right\}\]

Can you take things from here?