Solving an equation using Laplace Transforms

[Daniel Wood]

Hi guys. Im having a problem with one of my further maths assignments for college.
Unfortunately i am better with electronics than math. I hope someone would be able to provide some friendly advice on this question. it would be a great help

First i have to solve using laplace transforms

[LATEX]\frac{d^2x}{dt} -4x = 24cos(2t)[/LATEX]

given that t=0 and x = 3 and dx/dt = 4


I tried to have a dig at this myself and ended up getting stuck half way through

Firstly I written some shorthand to make solving this easier

[LATEX]\bar{x} = L \left\{ x \right\}[/LATEX]
[LATEX]x_0[/LATEX] = Value of x when t = 0
[LATEX]x_1[/LATEX] = dx/dt when t=0

So i took laplace transforms throughout and ended up with this
[LATEX]S^2 \bar{x}-Sx_0 -x_1 - 4 \bar{x} = \frac{24S}{S^2+4}[/LATEX]

Then i insert the initial condition's and rearrange the formula to get

[LATEX]\bar{x} = \frac{3S^3 +4S^2 + 36S +16}{(S^2 +4)(S-2)(S+2)} \equiv \frac {AS+B} {S^2+4} + \frac {C} {S - 2} + \frac {D} {S-2}[/LATEX]

Im quite sure it is correct up to this point. Anyway to find the value for A, B, C, D i would have to re-arrange the formula again to get

EDIT: 4S was meant to be 4S^2
[LATEX]3S^3 + 4S^2 + 36S + 16 = (AS+B)(S-2)(S+2) + C(S+2)(S^2 + 4) + D(S-2)(S^2 + 4)[/LATEX]

From that, I should be able to find the values by using S to cancel the other letters out.
S = 2 gets me a value of C = 4
S = -2 gets me a value of D = 2
Working out B does not work out so well though. If i put S = 0, the answer would not be a whole number. And from that I cant work out what A is

Maybe there is another method of solving this equation. Any suggestions would be great

 

[MrAl]

Hi guys. Im having a problem with one of my further maths assignments for college.
Unfortunately i am better with electronics than math. I hope someone would be able to provide some friendly advice on this question. it would be a great help

First i have to solve using laplace transforms

[LATEX]\frac{d^2x}{dt} -4x = 24cos(2t)[/LATEX]

given that t=0 and x = 3 and dx/dt = 4


I tried to have a dig at this myself and ended up getting stuck half way through

Firstly I written some shorthand to make solving this easier

[LATEX]\bar{x} = L \left\{ x \right\}[/LATEX]
[LATEX]x_0[/LATEX] = Value of x when t = 0
[LATEX]x_1[/LATEX] = dx/dt when t=0

So i took laplace transforms throughout and ended up with this
[LATEX]S^2 \bar{x}-Sx_0 -x_1 - 4 \bar{x} = \frac{24S}{S^2+4}[/LATEX]

Then i insert the initial condition's and rearrange the formula to get

[LATEX]\bar{x} = \frac{3S^3 +4S^2 + 36S +16}{(S^2 +4)(S-2)(S+2)} \equiv \frac {AS+B} {S^2+4} + \frac {C} {S - 2} + \frac {D} {S-2}[/LATEX]

Im quite sure it is correct up to this point. Anyway to find the value for A, B, C, D i would have to re-arrange the formula again to get

[LATEX]3S^3 + 4S + 36S + 16 = (AS+B)(S-2)(S+2) + C(S+2)(S^2 + 4) + D(S-2)(S^2 + 4)[/LATEX]

From that, I should be able to find the values by using S to cancel the other letters out.
S = 2 gets me a value of C = 4
S = -2 gets me a value of D = 2
Working out B does not work out so well though. If i put S = 0, the answer would not be a whole number. And from that I cant work out what A is

Maybe there is another method of solving this equation. Any suggestions would be great

Hi Daniel,


There is another method, but i have a feeling you just need to go over it again the way you did do it. For example, your equation for x bar is ok on the left, but not ok on the right because you have two terms with s-2 in the denominator instead of only one. Still, it looks like you may have not carried that error into the final equation but check that again anyway.

The other way to do it is to multiply out your equation for x bar so that you have powers of s on both sides like:
n1*s^3+n2*s^2+n3*s+n4=n5*s^3+n6*s^2+n7*s+n8

where the n's might also contain the variables A, B, C, and D. You then equate like powers of s forming equations:
n1*s^3=n5*s^3
n2*s^2=n6*s^2
n3*s=n7*s
n4=n8

That gives you four equations in four unknowns, which you can solve simultaneously for the variables A, B, C, and D.

But doing it using your original method works too. Once you get the equations multiplied out correctly (assuming you dont have it correct right now, which you may very well have done ok already) you can then substitute the values you have ALREADY found (for C and D assuming they are correct) and come up with a simpler equation in s and the remaining variables A and B, then substitute s=0 and you'll get a reasonable result for another variable, then you can substitute that variable with it's value and you'll get an equation you can solve for the last variable. That will give you all four variables.

If you still have trouble doing this i'll post the entire procedure, but i suggest you just go over it again starting from your equation for x bar after changing one s-2 to s+2. Everything up to that point is just fine as you thought.

LATER:
I just went over your last equation and it is in fact correct. This means that something just went wrong when you substituted the trial values for s into that equation. Once you get C and D (as you did) you can substitute them into that equation to make it even simpler, then substitute s=0 to get the next variable (as you tried) and then substitute that, then solve for the last variable.

 

[Daniel Wood]

Hi MrAl,
Thank you for the reply!

Unfortunately re-doing this procedure throws the same problem. C and D are fine, but B always ends up as 1.33 no matter how carefully I try. I noticed i think i little typo in the last equation. Ill have to see the papers at home but im sure it was meant to be 3S^3 + 4S^2 . . . . . i don't know if that made any difference to you?

I think the problem lies with the x bar equation somewhere.. I actually know that A is supposed to equal -4 and B = 0 but its proving it to myself that seems to be the big issue. Thanks for the help.

 

[MrAl]

Hi Daniel,


Yes im sorry i didnt mention that, but yes that equation should read:
3*s^3+4*s^2+36*s+16=(s-2)*(s^2+4)*D+(s+2)*(s^2+4)*C+(s-2)*(s+2)*(B+s*A)

and the difference is in that term on the left with the '4', it should be 4*s^2 not 4*s.

Evaluating that new equation with s=2 we get:
128=32*C
so
C=4

and evaluating with s=-2 we get:
-64=-32*D
so
D=2

Now evaluating that equation with those found values C=4 and D=2, we get:
3*s^3+4*s^2+36*s+16=(s-2)*(s+2)*(B+s*A)+4*(s+2)*(s^2+4)+2*(s-2)*(s^2+4)

Evaluating that with s=0 we get:
16=16-4*B
which means:
B=0

Evaluating the second equation with B=0 we get:
3*s^3+4*s^2+36*s+16=s^3*A-4*s*A+6*s^3+4*s^2+24*s+16

and evaluating this last equation with s=1 we get:
59=50-3*A
which means:
A=-3

So the values found are:
A=-3,
B=0,
C=4,
D=2

I've double checked this using two different methods so that's three methods that produce the same result, so these values MUST be correct. But we'll check our result anyway.

To verify, we insert the values into the right hand side of the original equation which is:
(A*s+B)/(s^2+4)+D/(s+2)+C/(s-2)

(and take note that D is over the s+2 and C is over the s-2, and this is very important too)

and we obtain:
(3*s^3+4*s^2+36*s+16)/((s^2+4)*(s-2)*(s+2))

which is the original equation so they must be correct.

So this means that your only problem was with the algebra. That is a common place to mess up an otherwise perfect result.

 

[Ian Rogers]

MrAl... I tried to PM you but your inbox is full and it was rejected......

 

[MrAl]

Hi,

Oh ok thanks for telling me, I cleared it out now.

 

[Daniel Wood]

Ahh I should have realised 0 multiplied by any multiple of B should be 0!
I realised exactly where I've gone wrong now.

Thank you for all your help