Second order differential help (Laplace)

In summary, the conversation discusses solving a second order differential equation using methods such as finding a particular solution and using Laplace transforms. There is also a minor mistake pointed out in one of the equations. The individual asking for help also shares their new approach to the problem.
  • #1
Bibbster
4
0
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
 

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  • #2
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.
 
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  • #3
Re: Second order differential help (laplace)

dwsmith said:
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 $d\cos(2t)+e\sin(2t)$.
 
  • #4
Re: Second order differential help (laplace)

dwsmith said:
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 :)
 
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  • #5
Re: Second order differential help (laplace)

dwsmith said:
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:

[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 I am thinking. (first term isn't split up as unsure ).

View attachment 1593
 

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  • #6
Re: Second order differential help (laplace)

Bibbster said:
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 I am thinking. (first term isn't split up as unsure ).

View attachment 1593

That fraction you are speaking of reduces to
\[
\frac{2}{s^2+4}-\frac{2}{s^2+s+4}
\]
 
  • #7
Re: Second order differential help (laplace)

dwsmith said:
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 I am struggling to tackle this one , if you could point me in right direction that'd be great
 
  • #8
Re: Second order differential help (laplace)

Bibbster said:
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 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?
 

FAQ: Second order differential help (Laplace)

What is a second order differential equation?

A second order differential equation is a mathematical equation that describes how a physical quantity changes in relation to another physical quantity, and involves the second derivative of the quantity being studied.

What is Laplace transform?

Laplace transform is a mathematical tool used to solve differential equations by transforming them from the time domain to the frequency domain. It is particularly useful for solving second order differential equations.

What is the Laplace transform of a second order differential equation?

The Laplace transform of a second order differential equation is a complex function that relates the input and output of a system in the frequency domain. It is represented as a ratio of polynomials in the Laplace variable s.

How is Laplace transform used to solve second order differential equations?

Laplace transform is used to transform a second order differential equation into an algebraic equation in the frequency domain, which can be solved using basic algebraic techniques. Once the solution is obtained, it can be transformed back to the time domain using the inverse Laplace transform.

What are the advantages of using Laplace transform to solve second order differential equations?

Some of the advantages of using Laplace transform to solve second order differential equations include its ability to solve complex equations, its simplicity in solving differential equations with varying initial conditions, and its usefulness in solving systems of differential equations.

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