i'm having trouble solving these two second order nonhomogenous DEs
y"+6y'+9y=(e^-3x)/(x^2+1)
and
(D^2+a^2)y = cosbt
where y'(0)=y(0)=0 and a and b are distict positive numbers
any kinda hints or help would be appreciated thanks
#1:
Variation of Parameters
From the auxilaiary equation we have $\displaystyle m^{2}+6m+9=0$
$\displaystyle C_{1}e^{-3x}+C_{2}xe^{-3x}$
From the Wronskian:
$\displaystyle W=\begin{vmatrix}e^{-3x}&xe^{-3x}\\-3e^{-3x}&(1-3x)e^{-3x}\end{vmatrix}=e^{-6x}$
$\displaystyle W_{1}=\begin{vmatrix}0&xe^{-3x}\\\frac{e^{-3x}}{1+x^{2}}&(1-3x)e^{-3x}\end{vmatrix}=\frac{-xe^{-6x}}{x^{2}+1}$
$\displaystyle W_{2}=\begin{vmatrix}e^{-3x}&0\\-3e^{-3x}&\frac{e^{-3x}}{1+x^{2}}=\frac{e^{-6x}}{1+x^{2}}$
$\displaystyle u'_{1}=\frac{W_{1}}{W}=\frac{-x}{x^{2}+1}$
$\displaystyle u'_{2}=\frac{W_{2}}{W}=\frac{1}{x^{2}+1}$
$\displaystyle u_{1}=\int\frac{-x}{x^{2}+1}dx=\frac{-ln(x^{2}+1)}{2}$
$\displaystyle u_{2}=\int\frac{1}{x^{2}+1}dx=tan^{-1}(x)$
So, we have:
$\displaystyle y=e^{-3x}\left(C_{1}+C_{2}x+xtan^{-1}(x)-\frac{ln(x^{2}+1)}{2}\right)$
The general solution is the sum of the general solution of theOriginally Posted by action259
homogeneous equation:
$\displaystyle
y''+a^2y=0
$,
and a particular integral of the inhomogeneous equation.
The solution of the homogeneous equation is routine, for the inhomogeneous
use a trial solution of the form:
$\displaystyle
y(t)=A \cos(bt) + B \sin(bt)
$
Then we find:
$\displaystyle
-A b^2 \cos(bt)-B b^2 \sin(bt)+$$\displaystyle A a^2 A \cos(bt)+B a^2 \sin(bt)=\cos(bT)
$,
so $\displaystyle B=0$, and $\displaystyle A=1/(a^2-b^2)$.
Fitting the initial conditions is now just routine and left to the reader.
RonL
Its a typo, you will note it is implicit in the solution given. Comes fromOriginally Posted by ThePerfectHacker
typing a TeX version of the answer off of a scrap of paper.
It will be corrected.
RonL
Yes, but I only get the computer for short periods as the kids are playingOriginally Posted by ThePerfectHacker
games on it all day at the moment. So I have to do as much as possible off
line!
RonL
(I keeps saying I will set up a wirless network, but every time I
get close, someone finds something else to spend the budget on )
I had been contemplating posting a variation of parameters solution, butOriginally Posted by galactus
there was so much TeX involved, and it looked so ugly I went off looking
for another method.
Looking at the final result it is so ugly this could only have been set as an
exercises in variation of parameters method.
RonL
(Any opinions expressed in this post are the authors and do not necessarily
reflect those of the management )