Welcome to our community

Be a part of something great, join today!

Second order differential help (Laplace)

Bibbster

New member
Oct 28, 2013
4
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 ..


image.jpg
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.

[HR][/HR]
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.
 
Last edited:

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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

New member
Oct 28, 2013
4
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.

[HR][/HR]
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 :)
 
Last edited:

Bibbster

New member
Oct 28, 2013
4
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.

[HR][/HR]
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:

[tex]\displaystyle \dfrac{2s}{(s^2+4)(s^2+s+4)} \ = \ \dfrac{2s}{(s^2+4)} + \dfrac{2s}{(s^2+s+4)} [/tex]

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: [tex]\displaystyle \ \dfrac{-s}{(s^2+s+4)} + \dfrac{1}{(s^2+s+4)} [/tex] which would then leave me to find possible solutions for [tex] \displaystyle (s^2+s+4) [/tex] with one being [tex] \displaystyle =(s+0.5)^2+15/4 [/tex]. Then after that (if correct) i'm once again confused . :(

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

maths help.JPG
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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:

[tex]\displaystyle \dfrac{2s}{(s^2+4)(s^2+s+4)} \ = \ \dfrac{2s}{(s^2+4)} + \dfrac{2s}{(s^2+s+4)} [/tex]

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: [tex]\displaystyle \ \dfrac{-s}{(s^2+s+4)} + \dfrac{1}{(s^2+s+4)} [/tex] which would then leave me to find possible solutions for [tex] \displaystyle (s^2+s+4) [/tex] with one being [tex] \displaystyle =(s+0.5)^2+15/4 [/tex]. Then after that (if correct) i'm once again confused . :(

below are my new workings to keep you guys on track with what im 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

New member
Oct 28, 2013
4
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 [tex]\dfrac{2}{(s^2+4)}[/tex][tex]-[/tex][tex]\dfrac{2}{(s^2+s+4)}[/tex] with [tex]\dfrac{2}{(s^2+4)}[/tex] becoming [tex]Sin2t[/tex] and [tex]\dfrac{2}{(s^2+s+4)}[/tex] becoming [tex]\dfrac{2}{(s^2+0.5)^2+15/4}[/tex] however im struggling to tackle this one , if you could point me in right direction that'd be great
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Re: Second order differential help (laplace)

as you said , i now have [tex]\dfrac{2}{(s^2+4)}[/tex][tex]-[/tex][tex]\dfrac{2}{(s^2+s+4)}[/tex] with [tex]\dfrac{2}{(s^2+4)}[/tex] becoming [tex]Sin2t[/tex] and [tex]\dfrac{2}{(s^2+s+4)}[/tex] becoming [tex]\dfrac{2}{(s^2+0.5)^2+15/4}[/tex] however im 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?