Laplace Transform Solution to Second Order ODE IVP

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Homework Help Overview

The discussion revolves around solving a second-order ordinary differential equation (ODE) with initial value conditions using the Laplace transform. The specific equation is y'' + 6y = f(t), where f(t) is defined piecewise. The participants are exploring the application of Laplace transforms to both the differential equation and the piecewise function.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the Laplace transform to the ODE and the piecewise function, questioning the algebraic manipulations and the correctness of the resulting expressions. There are considerations about the inverse Laplace transforms of various terms, particularly those involving exponential factors and unit step functions.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's algebra and reasoning. Some guidance has been offered regarding the use of partial fractions and the interpretation of time shifts in the context of the Laplace transform. There is no explicit consensus yet, as participants continue to clarify and correct each other's work.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of direct solutions provided. There are also discussions about potential algebraic errors and the need for careful attention to the definitions and transformations involved.

tetrakis
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Homework Statement


y''+6y=f(t), y(0)=0, y'(0)=-2

f(t)= t for 0≤t<1 and 0 for t≥1

Homework Equations




The Attempt at a Solution



L{y''}+6L{y}=L{t}-L{tμ(t-1)} where μ(t-1) is Unit Step

Y(s)=L{y}
sY(s)-y(0)=L{y'} and y(0)=0
s2Y(s)-sy(0)-y'(0)+6Y(s) where y(0)=0 and y'(0)=-2

LHS:
s2Y(s)+2+6Y(s)
Y(s)(s2+6)-2

RHS:
L{t}-L{tμ(t-1)}
L{t}=1/s2
L{tμ(t-1)}=L{(t-1)μ(t-1)+L{μ(t-1)}=e-s/s2+e-s/s
L{t}-L{tμ(t-1)}=1/s2-e-s/s2-e-s/s

Y(s)= -1/(s2+6) -e-s/(s2(s2+6))-e-s/s(s2+6)

taking the inverse laplace of the first term
-L-1{1/(s2+6)}=(-1/6)L-1{1/s}+(1/6)L-1{1/(s+6)}=(-1/6)+1/6e-6t)

how could I take the laplace of the other 2 terms?
I was thinking L-1{e-s/(s2(s2+6))}=y(t-1)μ(t-1)?

or I could just break it up into (1/6)L-1{1/s2}-(1/6)L-1{1/(s2+6)} which I know the inverse laplace transforms for, however when I do so I get
y(t)= (1/6)t-(1/6)+(1/6)e-6t
which seems a bit bizarre to me, I would have assumed that the answer would have a unit step function

Thank you for your time
 
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tetrakis said:

Homework Statement


y''+6y=f(t), y(0)=0, y'(0)=-2

f(t)= t for 0≤t<1 and 0 for t≥1

Homework Equations

The Attempt at a Solution



L{y''}+6L{y}=L{t}-L{tμ(t-1)} where μ(t-1) is Unit Step

Y(s)=L{y}
sY(s)-y(0)=L{y'} and y(0)=0
s2Y(s)-sy(0)-y'(0)+6Y(s) where y(0)=0 and y'(0)=-2

LHS:
s2Y(s)+2+6Y(s)
Y(s)(s2+6)-2

RHS:
L{t}-L{tμ(t-1)}
L{t}=1/s2
L{tμ(t-1)}=L{(t-1)μ(t-1)+L{μ(t-1)}=e-s/s2+e-s/s
L{t}-L{tμ(t-1)}=1/s2-e-s/s2-e-s/s
I think you're fine up to here.

Y(s)= -1/(s2+6) -e-s/(s2(s2+6))-e-s/s(s2+6)
You didn't do the algebra correctly. The first term is wrong.

taking the inverse laplace of the first term
-L-1{1/(s2+6)}=(-1/6)L-1{1/s}+(1/6)L-1{1/(s+6)}=(-1/6)+1/6e-6t)
You didn't factor the denominator correctly. ##s^2 + 6 \ne s(s+6)##. Check your table for a Laplace transform with a quadratic denominator.

how could I take the laplace of the other 2 terms?
I was thinking L-1{e-s/(s2(s2+6))}=y(t-1)μ(t-1)?

or I could just break it up into (1/6)L-1{1/s2}-(1/6)L-1{1/(s2+6)} which I know the inverse laplace transforms for, however when I do so I get
y(t)= (1/6)t-(1/6)+(1/6)e-6t
which seems a bit bizarre to me, I would have assumed that the answer would have a unit step function

Thank you for your time
 
Last edited:
vela said:
I think you're fine up to here.


What happened to the -2 from the LHS?


You didn't factor the denominator correctly. ##s^2 + 6 \ne s(s+6)##. Check your table for a Laplace transform with a quadratic denominator.

definitely an algebra error present,

Y(s)(s2+6)+2=1/s2-e-s/s2-e-s/s

then bring the 2 over
Y(s)(s2+6)=1/s2-e-s/s2-e-s/s-2

divided out the (s2+6) term

Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)
 
vela said:
I think you're fine up to here.


You didn't do the algebra correctly. The first term is wrong.


You didn't factor the denominator correctly. ##s^2 + 6 \ne s(s+6)##. Check your table for a Laplace transform with a quadratic denominator.


So, with the algebra corrected I have
Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)

the inverse of the first term
L-1{1/(s2(s2+6))} I can't find anything like this in my table, most of the entries have s2+k2 which I can't break 6 down into?
 
You have to use partial fractions on that term to break it up. If ##k^2=6##, then ##k = \sqrt{6}##, no?
 
vela said:
You have to use partial fractions on that term to break it up. If ##k^2=6##, then ##k = \sqrt{6}##, no?

right! so I have 1/(s2(s2+6))=(1/6)(1/s2) - (1/6)(1/(s2+6)

which the inverse laplace transform is

(1/6)t-(√6/6)sin(√6t) ?
 
Almost. You're off by a factor of ##\sqrt{6}## in the second term:
$$\frac{1}{6}\frac{1}{s^2+6} = \frac{1}{6\sqrt{6}} \frac{\sqrt{6}}{s^2+6} \rightarrow \frac{1}{6\sqrt{6}}\sin\sqrt6 t$$
 
vela said:
Almost. You're off by a factor of ##\sqrt{6}## in the second term:
$$\frac{1}{6}\frac{1}{s^2+6} = \frac{1}{6\sqrt{6}} \frac{\sqrt{6}}{s^2+6} \rightarrow \frac{1}{6\sqrt{6}}\sin\sqrt6 t$$

right, okay, and I can also apply this to the last term?

L-1{2/(s2+6}=(2/√6)sin(6t)?

how could I approach the exponential terms?
 
The exponential indicates a time shift. You found, for example, that
$$\frac{1}{s^2+6} \rightarrow \frac{1}{6}t - \frac{1}{6\sqrt{6}} \sin \sqrt{6}t,$$ so
$$e^{-s}\frac{1}{s^2+6} \rightarrow \frac{1}{6}(t-1)u(t-1) - \frac{1}{6\sqrt{6}} \sin \sqrt{6}(t-1)\,u(t-1).$$
 
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  • #10
vela said:
The exponential indicates a time shift. You found, for example, that
$$\frac{1}{s^2+6} \rightarrow \frac{1}{6}t - \frac{1}{6\sqrt{6}} \sin \sqrt{6}t,$$ so
$$e^{-s}\frac{1}{s^2+6} \rightarrow \frac{1}{6}(t-1)u(t-1) - \frac{1}{6\sqrt{6}} \sin \sqrt{6}(t-1)\,u(t-1).$$

okay, so for

$$e^{-s}\frac{1}{s(s^2+6)} \rightarrow \frac{1}{6}e^{-s}\frac{1}{s} -\frac{1}{6}e^{-s}\frac{s}{s^{2}+6} \rightarrow \frac{1}{6}u(t-1) - \frac{1}{6}cosh(\sqrt{6})u(t-1)$$?
 
  • #11
making the entire thing

$$ \frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cosh(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t) $$
 
  • #12
Where did you get cosh from? It should be cos. The argument of that function is slightly incorrect. The rest of the solution looks good.
 
Last edited:
  • #13
vela said:
Where did you get cosh from? It should be cos. The argument of that function is slightly incorrect. The rest of the solution looks good.

okay, fixed.

$$ \frac{1}{6}t - \frac{1}{6\sqrt{6}}sin(\sqrt{6}t) - \frac{1}{6}(t-1)u(t-1) + \frac{1}{6\sqrt{6}}sin(\sqrt{6}(t-1))u(t-1) - \frac{1}{6}u(t-1) + \frac{1}{6}cos(\sqrt{6}t)u(t-1) - \frac{2}{\sqrt{6}}sin(\sqrt{6}t) $$

I'm sorry but I'm not sure what you mean by the argument? should it be t-1 instead of t?
 
  • #14
Yes, t-1 instead of t. Everywhere t appears in the unshifted function, you have to replace it by t-1.
 
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