Answer:
[tex]y(t)=\frac{3}{5}e^t-\frac{3}{5}cost 3t-\frac{1}{5} sin3t[/tex]
Step-by-step explanation:
We are given that
y'-y=2 sin3t,y(0)=0
We have to solve given differential equation using Laplace transform
We know that L(y'-y)=L(2sin3t)
[tex]sY(s)-y(0)-Y(s)= \frac{6}{s^2+9}[/tex]
[tex]L(sinat)=\frac{a}{s^2+a^2}[/tex]
[tex]Y(s)(s-1)=\frac{6}{s^2+9}[/tex]
[tex]Y(s)=\frac{6}{(s^2+9)(s-1)}[/tex]
Using partial fraction
[tex]\frac{6}{(s^2+9)(s-1)}=\frac{A}{s-1}+\frac{Bs+c}{s^2+9}[/tex]
[tex]6=A(s^2+9)+(s-1)(Bs+C)[/tex]
s-1=0 then s=1
substitute s= 1 in equation
6=A(1+9)
[tex]A=\frac{6}{10}=\frac{3}{5}[/tex]
Comparing coefficient of [tex]s^2[/tex] and s on both sides then we get
[tex]A+B=0[/tex]
[tex]B=-A=-\frac{3}{5}[/tex]
[tex]-B+C=0[/tex]
[tex]B=C=-\frac{3}{5}[/tex]
Substitute the values
Then, Y(s)=[tex]\frac{3}{5}\frac{1}{s-1}+\frac{\frac{-3}{5} s-\frac{3}{5}}{s^2+9}[/tex]
Apply inverse transform then we get
[tex]y(t)=\frac{3}{5}e^t-\frac{3}{5}cost 3t-\frac{1}{5} sin3t[/tex]
[tex]L(cos at}=\frac{s}{s^+a^2}[/tex] and [tex]L(e^{at})=\frac{1}{s-a}[/tex]
