Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2, y'(0) 1

Answer:The solution of the differential equation is [tex]y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)[/tex]Step-by-step explanation:The differential equation is given by: y" + y = Sin(2t)i) Using characteristic equation:The characteristic equation method assumes that y(t)=[tex]e^{rt}[/tex], where "r" is a constant.We find the solution of the homogeneus differential equation:y" + y = 0[tex]y'=re^{rt}[/tex][tex]y"=r^{2}e^{rt}[/tex][tex]r^{2}e^{rt}+e^{rt}=0[/tex][tex](r^{2}+1)e^{rt}=0[/tex]As [tex]e^{rt}[/tex] could never be zero, the term (r²+1) must be zero:(r²+1)=0r=±iThe solution of the homogeneus differential equation is:[tex]y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}[/tex]Using Euler's formula:[tex]y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)][/tex][tex]y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)[/tex][tex]y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)[/tex]The particular solution of the differential equation is given by:[tex]y(t)_{p}=ASin(2t)+BCos(2t)[/tex][tex]y'(t)_{p}=2ACos(2t)-2BSin(2t)[/tex][tex]y''(t)_{p}=-4ASin(2t)-4BCos(2t)[/tex]So we use these derivatives in the differential equation:[tex]-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)[/tex][tex]-3ASin(2t)-3BCos(2t)=Sin(2t)[/tex]As there is not a term for Cos(2t), B is equal to 0.So the value A=-1/3The solution is the sum of the particular function and the homogeneous function:[tex]y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)[/tex]Using the initial conditions we can check that C1=5/3 and C2=2ii) Using Laplace Transform: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-1ℒ[y]=Y(s) ℒ[Sin(2t)]=[tex]\frac{2}{(s^{2}+4)}[/tex]We replace the previous data in the equation:s²·Y(s) -2s-1+Y(s) =[tex]\frac{2}{(s^{2}+4)}[/tex](s²+1)·Y(s)-2s-1=[tex]\frac{2}{(s^{2}+4)}[/tex](s²+1)·Y(s)=[tex]\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}[/tex]Y(s)=[tex]\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}[/tex]Y(s)=[tex]\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}[/tex]Using partial franction method:[tex]\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}[/tex]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=2B+D=1A+4C=8B+4D=6The solutions are:A=0 ; B= -2/3 ; C=2 ; D=5/3So,Y(s)=[tex]\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}[/tex]Y(s)=[tex]-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}[/tex]By using the inverse of the Laplace transform:ℒ⁻¹[Y(s)]=ℒ⁻¹[[tex]-\frac{1}{3} \frac{2}{s^{2}+4}[/tex]]-ℒ⁻¹[[tex]2\frac{s }{s^{2}+1}[/tex]]+ℒ⁻¹[[tex]\frac{5}{3}\frac{1}{s^{2}+1}[/tex]][tex]y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)[/tex]