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.