Laplace Transform Solution to Second Order ODE IVP

In summary: Thanks for your help!In summary, the solution to the given differential equation is given by y(t) = (1/6)t - (1/6√6)sin(√6t) - (1/6)(t-1)u(t-1) + (1/6√6)sin(√6(t-1))u(t-1) - (1/6)u(t-1) + (1/6)cos(√6t)u(t-1) - (2/√6)sin(√6t).
  • #1
tetrakis
14
0

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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  • #2
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:
  • #3
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)
 
  • #4
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?
 
  • #5
You have to use partial fractions on that term to break it up. If ##k^2=6##, then ##k = \sqrt{6}##, no?
 
  • #6
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) ?
 
  • #7
Almost. You're off by a factor of ##\sqrt{6}## in the second term:
$$\frac{1}{6}\frac{1}{s^2+6} = \frac{1}{6\sqrt{6}} \frac{\sqrt{6}}{s^2+6} \rightarrow \frac{1}{6\sqrt{6}}\sin\sqrt6 t$$
 
  • #8
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?
 
  • #9
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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Related to Laplace Transform Solution to Second Order ODE IVP

What is the Laplace Transform Solution to Second Order ODE IVP?

The Laplace Transform Solution to Second Order ODE IVP is a mathematical technique used to solve second order ordinary differential equations (ODEs) with initial conditions. It involves transforming the ODE from the time domain to the frequency domain using the Laplace transform, and then solving for the transformed function. After that, the inverse Laplace transform is applied to the solution to obtain the solution in the time domain.

When is the Laplace Transform Solution to Second Order ODE IVP used?

The Laplace Transform Solution to Second Order ODE IVP is typically used when solving ODEs with complicated initial conditions, or when the ODE cannot be solved using traditional methods. It is also useful for finding closed-form solutions to ODEs, as it can often simplify the problem and make it easier to solve.

What are the benefits of using the Laplace Transform Solution to Second Order ODE IVP?

The Laplace Transform Solution to Second Order ODE IVP has several benefits, including the ability to solve ODEs with complicated initial conditions, the ability to find closed-form solutions to ODEs, and the ability to simplify the problem and make it easier to solve. It is also a powerful tool for solving ODEs that cannot be solved using traditional methods.

What are the limitations of the Laplace Transform Solution to Second Order ODE IVP?

While the Laplace Transform Solution to Second Order ODE IVP has many benefits, it also has some limitations. It can only be used for linear ODEs, and the initial conditions must be known. It also requires knowledge of the Laplace transform and inverse Laplace transform, which can be complex and time-consuming to calculate.

Can the Laplace Transform Solution to Second Order ODE IVP be used for higher order ODEs?

Yes, the Laplace Transform Solution can be extended to solve higher order ODEs with initial conditions. However, the process becomes more complex and time-consuming, as more transforms and inverse transforms are required. It is more commonly used for second order ODEs, but can be applied to higher order ODEs if needed.

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