Using Laplace transform we have:L(x')+7L(x) = 5L(cos(2t))sL(x)-x(0) + 7L(x) = 5s/(s^2+4)(s+7)L(x)- 4 = 5s/(s^2+4)(s+7)L(x) = (5s - 4s^2 -16)/(s^2+4)
=> L(x) = -(4s^2 - 5s +16)/(s^2+4)(s+7)
now the boring part, using partial fractions we separate 1/(s^2+4)(s+7) that is:(7-s)/[53(s^2+4)] + 1/53(s+7). So:
L(x)= (1/53)[(-28s^2+4s^3-4s^2+35s-5s^2+5s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]L(x)= (1/53)[(4s^3 -37s^2 +40s)/(s^2+4) + (-4s^2+5s-16)/(s+7)]
denoting T:= L^(-1)and x= (4/53) T(s^3/(s^2+4)) - (37/53)T(s^2/(s^2+4)) +(40/53) T(s^2+4)-(4/53) T(s^2/s+7) +(5/53)T(s/s+7) - (16/53) T(1/s+7)
Try comparing your solution with the following:
Solution:
Answer:
Check:
![2[10-13(\frac{40}{17})]+9(\frac{40}{17})=-34(\frac{40}{17})+60\\-20=-20](https://tex.z-dn.net/?f=2%5B10-13%28%5Cfrac%7B40%7D%7B17%7D%29%5D%2B9%28%5Cfrac%7B40%7D%7B17%7D%29%3D-34%28%5Cfrac%7B40%7D%7B17%7D%29%2B60%5C%5C-20%3D-20)
<em>Hope this was helpful.</em>
Answer:
The answer to your question is x = 6 and y = 9
Step-by-step explanation:
Data
AB = x DF = 18
BC = 5 FE = 15
AC = 3 DE = y
Process
1.- Find y
Use proportions to find it
DE/FE = AC/BC
Substitution
y/15 = 3/5
Solve for y
y = 15(3/5)
y = 45/5
y = 9
2.- Find x
AB/DF = BC/FE
Substitution
x/18 = 5/15
Solve for x
x = 18(5/15)
x = 90/15
x = 6
C(x) = 1200(40) + 100
or
= mx+c
where m = 1200, x = 40, y = 100
hope it helped,
click thx if it did.
