Question

Use the Laplace transformation to solve the given IVP. y'-y=2sin3t, y(0)=0

Answer 1

Answer:

$y(t)=\frac{3}{5}e^t-\frac{3}{5}cost 3t-\frac{1}{5} sin3t$

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)

$sY(s)-y(0)-Y(s)= \frac{6}{s^2+9}$

$L(sinat)=\frac{a}{s^2+a^2}$

$Y(s)(s-1)=\frac{6}{s^2+9}$

$Y(s)=\frac{6}{(s^2+9)(s-1)}$

Using partial fraction

$\frac{6}{(s^2+9)(s-1)}=\frac{A}{s-1}+\frac{Bs+c}{s^2+9}$

$6=A(s^2+9)+(s-1)(Bs+C)$

s-1=0 then s=1

substitute s= 1 in equation

6=A(1+9)

$A=\frac{6}{10}=\frac{3}{5}$

Comparing coefficient of $s^2$ and s on both sides then we get

$A+B=0$

$B=-A=-\frac{3}{5}$

$-B+C=0$

$B=C=-\frac{3}{5}$

Substitute the values

Then, Y(s)=$\frac{3}{5}\frac{1}{s-1}+\frac{\frac{-3}{5} s-\frac{3}{5}}{s^2+9}$

Apply inverse transform then we get

$y(t)=\frac{3}{5}e^t-\frac{3}{5}cost 3t-\frac{1}{5} sin3t$

$L(cos at}=\frac{s}{s^+a^2}$ and $L(e^{at})=\frac{1}{s-a}$