1. ## second order DEs

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

2. #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:

$\displaystyle y=e^{-3x}\left(C_{1}+C_{2}x+xtan^{-1}(x)-\frac{ln(x^{2}+1)}{2}\right)$

3. Originally Posted by action259
(D^2+a^2)y = cosbt
where y'(0)=y(0)=0 and a and b are distict positive numbers
The general solution is the sum of the general solution of the
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 4. Originally Posted by CaptainBlank Then we find: \displaystyle -A b^2 \cos(bt)-B b^2 \sin(bt)+a^2 \cos(bt)=\cos(bT) , I belive you made a mistake. You did not substitute after the \displaystyle a^2. 5. Originally Posted by ThePerfectHacker I belive you made a mistake. You did not substitute after the \displaystyle a^2. Its a typo, you will note it is implicit in the solution given. Comes from typing a TeX version of the answer off of a scrap of paper. It will be corrected. RonL 6. Originally Posted by CaptainBlack Comes from typing a TeX version of the answer off of a scrap of paper. I do not do that. I simply type do problem through LaTeX. Sometimes I spend much time because I am unable to complete the solution in the end and delete my post. 7. Originally Posted by ThePerfectHacker I do not do that. I simply type do problem through LaTeX. Sometimes I spend much time because I am unable to complete the solution in the end and delete my post. Yes, but I only get the computer for short periods as the kids are playing 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 ) 8. Originally Posted by galactus #1: Variation of Parameters I had been contemplating posting a variation of parameters solution, but 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 ) 9. Originally Posted by CaptainBlack 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)$
How can you find that ? When I put Acos(bt) + Bsin(bt) in the original equation I get 0 = cos(bt), how can you have found A and B with it ?