Homework Help, Solving particular solution for ODE (a)Find the particular solution to 
$y''+2y'+y=\frac{5.5e^{-t}}{t^2+1}$
my question is how to find particular solution when right hand side is negative power
(b)$x^2y''+19xy'+81y=x^7$
by Euler's equation, let $t=lnx$
we have $\frac{d^2y}{dt^2}+18\frac{dy}{dt}+81y$ but this is only the case when its homogeneous equation. I don't know how the right hand side would be after the substitution
 A: Hint:
$$y'' + 2y' + y = e^{-t}(y e^{t})'' = \frac{5.5e^{-t}}{(1+t^2)}$$
Now multiply both sides by $e^{t}$ and integrate twice. Two useful integrals are: $$\int \frac{dt}{1+t^2} = \arctan(t)$$
$$\int\arctan(t)dt = t\arctan(t) - \frac{1}{2}\log(1+t^2)$$
A: I think Winther's got a good answer to a.  As for b, you have $t=\ln x$, so $x=e^t$.  So the right side of the equation
$$x^7=(e^t)^7=e^{7t}$$
A: For your particular solution in question 2,  note that we want to end up with a function that's $x^7$,  and and we have $x^2y''+19xy'+81y$ on the right hand side.  This naturally suggests that your solution should be of the form $y=cx^7$ for some constant c,  because if you take $y'$ and $y''$ each time you lose a power of x but then make it back up with the coefficient.
Can you take it from here?  If not, I can continue.
So, this is called the method of undetermined coeffficients, where you can figure out the form your answer should take and then find the coefficients that make it true.
so, taking a solution of the form $y=cx^7$ for some appropriate constant $c$,  we have $y'=7cx^6$,$y''=42cx^5$.  Now, since we are assuming this is a solution for our original differential equation, that means when I plug in these values of $y,y',y''$, I should get the right hand side.
So,  plugging in,  I get $x^2(42cx^5)+19x(7cx^6)+81cx^7=x^7$
Combining all the coefficients on the left hand side and factoring out the $x^7$, we are left with $(42c+323c+81c)x^7=x^7$.  Now, two polynomials are equal if and only if the coeffiecents are equal, so we get $42c+323c+81c=1$.   adding up those terms gives us $446c=1$,  hence $c=\frac 1 {446}$   Hence our particular solution is $y_p =\frac 1 {446} x^7$ (assuming I didn't make any arithmetic errors in this, of course) 
(and do the homogeneous in the normal way)
