Variation of parameters: $3y''+4y'+4=(\sin(t))e^{-t}$ Solve the following initial-value problem:
$$3y''+4y'+4=(\sin(t))e^{-t}$$
where $y(0)=1, y'(0)=0$
Here are my steps:
I started out with the homogeneous equation:
$$3y''+4y'+4=0$$
and found the roots to be $\frac{-1}{3}$ and $-1$
So that means two of the solutions are:
$y_1(t)=e^{\frac{-t}{3}}$
$y_2(t)=e^{-t}$
I was about to use the wronskian but then I just decided to overlook my book and realize that the author put a note saying:
How to make a note that the coefficient of $y''$ is 3. How will that affect me?
 A: Simply divide through by $3$ to bring the equation into standard form, e.g.
$$y''+\frac{4}{3}y+ \frac{4}{3} = 0$$
with corresponding characteristic equation $r^2 + \frac{4}{3}r + \frac{4}{3}=0$.
Note that although this does not change the roots of the characteristic equation, it does alter the $g(x)$ on the right hand side of the non-homogeneous equation (for this case, $g(x)=\frac{e^{-t}\sin t}{3}$), and so your formulas for variation of parameters won't work unless the DE is in standard form.
To compute a particular solution of the non-homogeneous system, determine $W(y_1, y_2)$ and use $Y(t)= u_1 y_1 + u_2 y_2$ where $y_1$ and $y_2$ are the solutions to the homogeneous DEs found in your post, and 
$$u_1(x) = -\int \frac{y_1(x) g(x)}{W(y_1,y_2)} \, dx, \quad 
u_2(x) = \int \frac{y_2(x) g(x)}{W(y_1,y_2)} \, dx.$$
A: let us make a change of variable $$ye^t = u,y = ue^{-t}, y' = u'e^{-t} - ue^{-t}, y'' = u''e^{-t}-2u'e^{-t}+ue^{-t} $$
so that $$(\sin(t))e^{-t} = 3y''+4y'+4y= \left( 3(u'' - 2u' + u) + 4(u'-u)+4u\right)e^{-t} $$ that is $u$ satisfies $$3u''-2u'+3u = \sin t $$ 
since you have $u'' + u = 0$ a particular solution is $$ u =\frac 12 \cos t   , y = \frac 1 2 e^{-t}\cos t $$
try to find $a, b$ so that $$y = ae^{-t} + be^{-t/3} +  \frac 1 2 e^{-t}\cos t$$ satisfies the initial conditions.
