Initial conditions
Please Help
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.
@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
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
You fraction is:
As usual, multiply on both sides by the denominator on the left to get
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.