The equation can be solved without using Laplace Transform.
Write derivatives of y as y' and y".
y"+4y=sin(x).
Let y=y₁+y₂ so y'=y₁'+y₂' and y"=y₁"+y₂".
If y₁"+4y₁=0, then the general solution is y₁=Asin(2x)+Bcos(2x), where A and B are constants to be found from initial conditions.
If y₂"+4y₂=sin(x) we can surmise that y₂=Csin(x)+Dcos(x), where C and D are constants to be found.
y₂'=Ccos(x)-Dsin(x) and y₂"=-Csin(x)-Dcos(x).
Therefore y₂"+4y₂=-Csin(x)-Dcos(x)+4Csin(x)+4Dcos(x)=
3Csin(x)+3Dcos(x)=sin(x).
So, matching coefficients, D=0 and C=⅓, y₂=⅓sin(x).
y=y₁+y₂=Asin(2x)+Bcos(2x)+⅓sin(x).
y(0)=2=B, y'=2Acos(2x)-2Bsin(2x)+⅓cos(x).
y'(0)=-1=2A+⅓, 2A=-4/3 and A=-⅔.
Therefore y=-⅔sin(2x)+2cos(2x)+⅓sin(x).
Using Laplace Transform:
ℒ{y"}=s²Y(s)-sy(0)-y'(0)=s²Y(s)-2s+1.
ℒ{y"+4y}=ℒ{sin(x)},
s²Y(s)-2s+1+4Y(s)=1/(1+s²),
Y(s)(s²+4)=1/(1+s²)+2s-1=(1+(2s-1)(1+s²))/(1+s²),
Y(s)=(2s³-s²+2s)/((1+s²)(4+s²)).
We need to write the right hand side as partial fractions:
(as+b)/(1+s²)+(cs+d)/(4+s²)≡(2s³-s²+2s)/((1+s²)(4+s²)), where a, b, c, d are constants.
Therefore:
(as+b)(s²+4)+(cs+d)(s²+1)≡(2s³-s²+2s)/((1+s²)(4+s²)).
as³+bs²+4as+4b+
cs³+ds²+ cs+ d=
2s³- s²+ 2s+ 0
From this:
①a+c=2
②b+d=-1
③4a+c=2
④4b+d=0
③-① gives us 3a=0, so a=0 and c=2;
④-② gives us 3b=1, b=⅓, d=-⁴⁄₃.
Y(s)=⅓/(1+s²)+(2s-⁴⁄₃)/(4+s²),
Y(s)=⅓/(1+s²)+2s/(4+s²)-⅔2/(4+s²).
This has been written so as to make inverse Laplace easy:
y(x)=⅓sin(x)+2cos(2x)-⅔sin(2x).
The solutions concur.