Find general solution of differential eqn

[DryRun]

Homework Statement
y''+4y'+3y=\sinh x, \, y(0)=y'(0)=0
The attempt at a solution
The auxiliary equation is: m^2+4m+3=0
m= -3 and -1
The complementary function is: y_c=Ae^{-3x}+Be^{-x}

I need to find the particular integral next. But i don't know what is the standard inverse D operator result in this case, but I'm going to deduce it from what information i do know.

L(D^2)\sinh ax=L(a^2)\sinh ax
From this, i will deduce (maybe I'm wrong) the principle \frac{1}{L(D^2)}\sinh ax=\frac{1}{L(a^2)}\sinh ax, where L(a^2)\not =0
L(D)=D^2+4D+3
a=1, k=1
y_p=\frac{1}{1+4D+3}\sinh x=\frac{1}{4D+4}\sinh x
On multiplying numerator and denominator by (4D-4), \frac{4D-4}{16D^2-16}\sinh x
And then i applied the same principle again: \frac{4D-4}{0}\sinh x

And then i got stuck, since the denominator is zero. I must have done something wrong somewhere but i have no idea where.

 

[HallsofIvy]

Sinh x= (e^x- e^{-x})/2 so that your solution to the homogeneous equation, e^{-x} is part of the particular solution. Try a particular solution of the form Ae^x+ Bxe^{-x}.

 

[DryRun]

Unfortunately, i don't understand the method. I need a formula that i can always use whenever i see sinhx but it's unfortunately not listed in my notes.

For example, in my notes, \frac{1}{L(D)}ke^{ax} gives the generic answer: \frac{1}{L(a)}ke^{ax}

And also, in my notes, \frac{1}{L(D)}ksin(ax+b) gives the generic answer: \frac{1}{L(-a^2)}ksin(ax+b)

But there is nothing about \sinh x, so i deduced it myself from the above formula:
\frac{1}{L(D)}ksinh(ax+b) gives the generic answer: \frac{1}{L(a^2)}ksinh(ax+b) but somehow it's wrong as it doesn't solve the problem in post #1.

 

[DryRun]

I finally got it! HallsofIvy, I'm sorry for dismissing your suggestion without looking deeper into it. I converted the R.H.S. to \frac{e^x-e^{-x}}{2} which is the equivalent of \sinh x and the correct answer revealed itself. Thank you for your help.