Answer:
The solution of the differential equation is ![y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)](https://tex.z-dn.net/?f=y%28t%29%3D%20-%20%5Cfrac%7B1%7D%7B3%7D%20Sin%282t%29%2B2%20Cos%28t%29%2B%5Cfrac%7B5%7D%7B3%7D%20Sin%28t%29)
Step-by-step explanation:
The differential equation is given by: y" + y = Sin(2t)
<u>i) Using characteristic equation:</u>
The characteristic equation method assumes that y(t)=
, where "r" is a constant.
We find the solution of the homogeneus differential equation:
y" + y = 0
![y'=re^{rt}](https://tex.z-dn.net/?f=y%27%3Dre%5E%7Brt%7D)
![y"=r^{2}e^{rt}](https://tex.z-dn.net/?f=y%22%3Dr%5E%7B2%7De%5E%7Brt%7D)
![r^{2}e^{rt}+e^{rt}=0](https://tex.z-dn.net/?f=r%5E%7B2%7De%5E%7Brt%7D%2Be%5E%7Brt%7D%3D0)
![(r^{2}+1)e^{rt}=0](https://tex.z-dn.net/?f=%28r%5E%7B2%7D%2B1%29e%5E%7Brt%7D%3D0)
As
could never be zero, the term (r²+1) must be zero:
(r²+1)=0
r=±i
The solution of the homogeneus differential equation is:
![y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}](https://tex.z-dn.net/?f=y%28t%29_%7Bh%7D%3Dc_%7B1%7De%5E%7Bit%7D%2Bc_%7B2%7De%5E%7B-it%7D)
Using Euler's formula:
![y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]](https://tex.z-dn.net/?f=y%28t%29_%7Bh%7D%3Dc_%7B1%7D%5BSin%28t%29%2BiCos%28t%29%5D%2Bc_%7B2%7D%5BSin%28t%29-iCos%28t%29%5D)
![y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)](https://tex.z-dn.net/?f=y%28t%29_%7Bh%7D%3D%28c_%7B1%7D%2Bc_%7B2%7D%29Sin%28t%29%2B%28c_%7B1%7D-c_%7B2%7D%29iCos%28t%29)
![y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)](https://tex.z-dn.net/?f=y%28t%29_%7Bh%7D%3DC_%7B1%7DSin%28t%29%2BC_%7B2%7DCos%28t%29)
The particular solution of the differential equation is given by:
![y(t)_{p}=ASin(2t)+BCos(2t)](https://tex.z-dn.net/?f=y%28t%29_%7Bp%7D%3DASin%282t%29%2BBCos%282t%29)
![y'(t)_{p}=2ACos(2t)-2BSin(2t)](https://tex.z-dn.net/?f=y%27%28t%29_%7Bp%7D%3D2ACos%282t%29-2BSin%282t%29)
![y''(t)_{p}=-4ASin(2t)-4BCos(2t)](https://tex.z-dn.net/?f=y%27%27%28t%29_%7Bp%7D%3D-4ASin%282t%29-4BCos%282t%29)
So we use these derivatives in the differential equation:
![-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)](https://tex.z-dn.net/?f=-4ASin%282t%29-4BCos%282t%29%2BASin%282t%29%2BBCos%282t%29%3DSin%282t%29)
![-3ASin(2t)-3BCos(2t)=Sin(2t)](https://tex.z-dn.net/?f=-3ASin%282t%29-3BCos%282t%29%3DSin%282t%29)
As there is not a term for Cos(2t), B is equal to 0.
So the value A=-1/3
The solution is the sum of the particular function and the homogeneous function:
![y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)](https://tex.z-dn.net/?f=y%28t%29%3D%20-%20%5Cfrac%7B1%7D%7B3%7D%20Sin%282t%29%20%2B%20C_%7B1%7D%20Sin%28t%29%20%2B%20C_%7B2%7D%20Cos%28t%29)
Using the initial conditions we can check that C1=5/3 and C2=2
<u>ii) Using Laplace Transform:</u>
To solve the differential equation we use the Laplace transformation in both members:
ℒ[y" + y]=ℒ[Sin(2t)]
ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]
By using the Table of Laplace Transform we get:
ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s)
-2s-1
ℒ[y]=Y(s)
ℒ[Sin(2t)]=![\frac{2}{(s^{2}+4)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28s%5E%7B2%7D%2B4%29%7D)
We replace the previous data in the equation:
s²·Y(s)
-2s-1+Y(s)
=![\frac{2}{(s^{2}+4)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28s%5E%7B2%7D%2B4%29%7D)
(s²+1)·Y(s)-2s-1=![\frac{2}{(s^{2}+4)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28s%5E%7B2%7D%2B4%29%7D)
(s²+1)·Y(s)=![\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28s%5E%7B2%7D%2B4%29%7D%2B2s%2B1%3D%5Cfrac%7B2%2B2s%28s%5E%7B2%7D%2B4%29%2Bs%5E%7B2%7D%2B4%7D%7B%28s%5E%7B2%7D%2B4%29%7D)
Y(s)=![\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%2B2s%28s%5E%7B2%7D%2B4%29%2Bs%5E%7B2%7D%2B4%7D%7B%28s%5E%7B2%7D%2B4%29%28s%5E%7B2%7D%2B1%29%7D)
Y(s)=![\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}](https://tex.z-dn.net/?f=%5Cfrac%7B2s%5E%7B3%7D%2Bs%5E%7B2%7D%2B8s%2B6%7D%7B%28s%5E%7B2%7D%2B4%29%28s%5E%7B2%7D%2B1%29%7D)
Using partial franction method:
![\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}](https://tex.z-dn.net/?f=%5Cfrac%7B2s%5E%7B3%7D%2Bs%5E%7B2%7D%2B8s%2B6%7D%7B%28s%5E%7B2%7D%2B4%29%28s%5E%7B2%7D%2B1%29%7D%3D%5Cfrac%7BAs%2BB%7D%7Bs%5E%7B2%7D%2B4%7D%20%2B%5Cfrac%7BCs%2BD%7D%7Bs%5E%7B2%7D%2B1%7D)
2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)
2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)
We solve the equation system:
A+C=2
B+D=1
A+4C=8
B+4D=6
The solutions are:
A=0 ; B= -2/3 ; C=2 ; D=5/3
So,
Y(s)=![\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}](https://tex.z-dn.net/?f=%5Cfrac%7B-%5Cfrac%7B2%7D%7B3%7D%20%7D%7Bs%5E%7B2%7D%2B4%7D%20%2B%5Cfrac%7B2s%2B%5Cfrac%7B5%7D%7B3%7D%20%7D%7Bs%5E%7B2%7D%2B1%7D)
Y(s)=![-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%2B4%7D%20%2B2%5Cfrac%7Bs%20%7D%7Bs%5E%7B2%7D%2B1%7D%2B%5Cfrac%7B5%7D%7B3%7D%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%2B1%7D)
By using the inverse of the Laplace transform:
ℒ⁻¹[Y(s)]=ℒ⁻¹[
]-ℒ⁻¹[
]+ℒ⁻¹[
]
![y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)](https://tex.z-dn.net/?f=y%28t%29%3D%20-%20%5Cfrac%7B1%7D%7B3%7D%20Sin%282t%29%2B2%20Cos%28t%29%2B%5Cfrac%7B5%7D%7B3%7D%20Sin%28t%29)