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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:

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- #3

- Jan 26, 2012

- 995

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 $d\cos(2t)+e\sin(2t)$.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.

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- #4

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:

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- #5

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

[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 ).

That fraction you are speaking of reduces toI 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

\[

\frac{2}{s^2+4}-\frac{2}{s^2+s+4}

\]

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- #7

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 greatThat fraction you are speaking of reduces to

\[

\frac{2}{s^2+4}-\frac{2}{s^2+s+4}

\]

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- #8

- Jan 26, 2012

- 995

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

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?