Solving an ODE with the Laplace transform

In summary, the conversation discusses solving a differential equation with an initial condition and finding the inverse Laplace transform using two different methods. The first attempt is corrected to agree with the results from the second method. The conversation also addresses completing the square incorrectly and the number of poles and zeros in a transfer function.
  • #1
PainterGuy
940
69
Moved from a technical forum, so homework template missing
Hi again,

The previous problem was done using y′′(t)+2y′(t)+10y(t)=10 with with intial condition y(0⁻)=0.

In the following case, I'm using an initial condition and setting the right hand side equal to zero.

Find y(t) for the following differential equation with intial condition y(0⁻)=4.
y′′(t)+2y′(t)+10y(t)=0

Using the following formulas.
ℒy(t)=Y(s)
ℒ{y′(t)}=sY(s)-f(0⁻)=sY(s)-4
ℒ{y′′(t)}=ℒ{(sY(s)-4)′}=ℒ{(sY(s)′-4′)}=ℒ{(sY(s)′)}=s²Y(s)-4s

y′′(t)+2y′(t)+10y(t)=0
⇒{s²Y(s)-4s}+2{sY(s)-4}+10Y(s)=0
⇒s²Y(s)-4s+2sY(s)-8+10Y(s)=0
⇒s²Y(s)+2sY(s)+10Y(s)-4s-8=0
⇒s²Y(s)+2sY(s)+10Y(s)=4s+8
⇒Y(s){s²+2s+10}=4s+8
⇒Y(s) = (4s+8) / (s²+2s+10)

Now finding the inverse Laplace transform.

Method 1:
https://www.physicsforums.com/attac...261154/?hash=ab39a4f0474f3384a6a5c37cdc441e05

1587534843424.png


Method 2:

This method I found from this video.

1587534952010.png
1587534990407.png


Question:

You can see that I'm getting different functions, f(t), for the mentioned methods. Where am I going wrong? Which f(t) is correct? Thank you for the help.
 

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  • #3
In your first attempt, you complete the square incorrectly: ##s^2+2s+10 = (s+1)^2+9##.
 
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  • #4
Mark44 said:
For a 2nd-order DE, you need two initial conditions, typically f(0) and f'(0).
This formula from above isn't correct:

Thank you. Yes, sorry, I didn't realize it at that time.

The previous problem was done using y′′(t)+2y′(t)+10y(t)=10 with with intial condition y(0⁻)=0. If y(t) represents displacement then its first derivative would be velocity function, v(t). As an initial condition it is assumed that y'(0⁻)=v(0⁻)=0. Now we have two initial conditions, i.e. two pieces of information about the system just a moment before it's analytically observed.

1587593819767.png


vela said:
In your first attempt, you complete the square incorrectly: ##s^2+2s+10 = (s+1)^2+9##.

I have corrected Method 1 below. Now the results from both Methods agree. But I have a related question which is below the plot.

1587594029421.png


1587594066709.png


Question:
Apparently I completed the square wrongly but overall my completed square agrees with the original expression s²+2s+10. So, does completing a square method require that no fractions or decimals should be used, or what else is wrong with my method? Please guide me.

s²+2αs+α²+β²=s²+2s+10=s²+2(1/2)s+(1/2)²+(3.1225)²={s+(1/2)}²+3.1225²=(s+α)²+β²
 
  • #5
PainterGuy said:
s²+2(1/2)s+(1/2)²+(3.1225)²
Without assuming what the answer is, try simplifying that expression. I guarantee you do not get what you started with.
 
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  • #6
Let's try it: s²+2(1/2)s+(1/2)²+(3.1225)² =s²+(2/2)s+(1/4)+9.750=s²+s+0.25+9.750=s²+s+10

original expression: s²+2s+10

s²+2s+10 ≠ s²+s+10

Sorry! :sorry:
 
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  • #7
Hi,

I read somewhere that for a transfer function, the number of poles (roots of denominator) and zeros (roots of numerator) is equal. The transfer function from post #1, Y(s)=(4s+8) / (s²+2s+10), has poles at -1±3i and a zero at -2. Where is the second zero located? Thank you for the help!
 
  • #8
Could someone please comment on my previous post?
 
  • #9
Original expression: Y(s)=(4s+8) / (s²+2s+10)

I think we could divide both numerator and denominator by "s²".

(1/s²)(4s+8) / (1/s²)(s²+2s+10) = (4/s+8/s²) / (1+2/s+10/s²)

As s→∞ (4/s+8/s²) / (1+2/s+10/s²) = 0/1 = 0

I think it gives us a zero at infinity.
 

Related to Solving an ODE with the Laplace transform

1. What is the Laplace transform?

The Laplace transform is a mathematical operation that converts a function of time into a function of complex frequency. It is often used in engineering and physics to solve differential equations, as it transforms a differential equation into an algebraic equation that is easier to solve.

2. How is the Laplace transform used to solve ODEs?

The Laplace transform is used to solve ODEs by transforming the differential equation into an algebraic equation, which can then be solved using standard algebraic techniques. The inverse Laplace transform is then applied to the solution to obtain the original function in the time domain.

3. What are the advantages of using the Laplace transform to solve ODEs?

The Laplace transform has several advantages for solving ODEs. It can handle a wide range of initial conditions, it can be used to solve systems of differential equations, and it can also be used to solve non-linear equations. Additionally, the Laplace transform is a powerful tool for solving ODEs that do not have closed-form solutions.

4. Are there any limitations to using the Laplace transform for solving ODEs?

While the Laplace transform is a useful tool for solving ODEs, it does have some limitations. It can only be applied to linear, time-invariant systems, and it may not be suitable for solving certain types of boundary value problems. Additionally, the inverse Laplace transform can be challenging to compute for some functions.

5. Are there any alternative methods for solving ODEs besides the Laplace transform?

Yes, there are several alternative methods for solving ODEs, such as the method of undetermined coefficients, the variation of parameters method, and numerical methods such as Euler's method and Runge-Kutta methods. The choice of method depends on the specific problem and the desired level of accuracy.

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