# Find general solution of differential eqn

Gold Member
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.

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

Gold Member
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.

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Gold Member
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.