# Thread: Laplace Transforms of DE

1. ## Laplace Transforms of DE

$\displaystyle y'' + 10y' + 25y = cos4t$

Initial conditions
$\displaystyle y(0) = y'(0) = 0$

2. Originally Posted by khanim
y" + 10y' + 25y = cos4t

Initial conditions
y(0) = y'(0) = 0

Where exactly are you stuck?

For starters, you're expected to look up the required transforms in a table. Letting LT[y] = Y, can you use your tables to get:

1. LT[y''] and LT[y']?

2. LT[cos(4t)]?

Substitute into the DE and make Y the subject, that is, get Y as a function of s. Can you do this?

When you've done all this things, please show what you've got (all working, please) and where you're stuck. If you can't do these things, please show what you can do and state where you're stuck.

3. Indicating with the symbol $\displaystyle L \{f(*)\}$ ther Laplace transform of a function $\displaystyle f(*)$ we have...

$\displaystyle L\{y(t)\} = Y(s)$

$\displaystyle L \{y^{'} (t)\} = s Y(s) - y(0)$

$\displaystyle L \{y^{''} (t)\} = s^{2} Y(s) - s y(0) - y^{'} (0)$ (1)

... so that is...

$\displaystyle Y(s) (s^{2} + 10 s + 25) = \frac{s}{s^{2} + 16}$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. I can do that.
My problem arises when I have to use partial fractions

This is where I am stuck
L{y} = P/(p2 +16)(p+5)(p+5)

I can assign variables to the denominators and solve but don't get the correct answer

5. May be that you don't consider that the resulting complex function $\displaystyle Y(s)$ has a pole of order two in $\displaystyle s=-5$ and in this case the partial fraction reduction in 'standard form' doesn't work ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. @chisigma:
I have tried it both ways i.e.

L{y)} = s/ (s2+16)(s+5)2
= As + B / s2 + 16 + Cs + D/ (s + 5)2

and

L{y} = p/ (p2+16)(p+5)2
= Ap + B/ (p2+16) + Cp+D/(p+5)2

Still stuck

7. The correct partial fraction expansion is...

$\displaystyle Y(s)= \frac{s}{ (s^{2} + 16) (s+5)^{2}} = \frac{as+b}{s^{2} + 16} + \frac{c} {s+5} + \frac{d} {(s+5)^{2}}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. Chi Sigma thx for the help, my maths was never that good

Y(s)= \frac{s}{ (s^{2} + 16) (s+5)^{2}} = \frac{as+b}{s^{2} + 16} + \frac{c} {s+5} + \frac{d} {(s+5)^{2}}

I have tried it but it turns out to be a major calculation, whats the simplest way of solving this partial fraction

9. You fraction is:
$\displaystyle Y(s)= \frac{s}{ (s^{2} + 16) (s+5)^{2}} = \frac{as+b}{s^{2} + 16} + \frac{c} {s+5} + \frac{d} {(s+5)^{2}}$

As usual, multiply on both sides by the denominator on the left to get
$\displaystyle s= (as+b)(s+5)^2+ c(s+5)(s^2+ 16)+ d(s^2+ 16)$
One way to do this is to do all of the indicated multiplications on the left side, then set "corresponding coefficients" equal: the coefficient pf [tex]s^2] will be 0, of s will be 1 and the constant term will be 1.

Or, since this has 4 unknown numbers, and must be true for all s, choose any convenient values for s to get 4 equations for a, b, c, and d. "-5" is particularly convenient; with s= -5, it becomes -5= (-5a+ b)(0)+ c(0) d(41). So d= -5/41.

No number will make the other coeficients 0 so just choose simple numbers. With s= 0, 0= 25b+ 80c+ 16d. With s= 1, 1= 36(a+b)+102c+ 17d. With s= -1, -1= 15(b-a)+ 68c+ 17d.

10. Thx Guys :-)