# Inhomogeneous equation

• Apr 30th 2011, 10:13 AM
AkilMAI
Inhomogeneous equation
How can I solve this equation $\displaystyle x''-x=(1+e^{t})^{-1}$ using the method of variation of parameters?Using the auxiliary equation the roots are $\displaystyle r_{1}=0$ and$\displaystyle r_{2}=1$.
I don't know if I'm suppose to look for a solution of the form $\displaystyle x(t)=p+q*e^{x}$ manly because the initial equation has o 1+... after the equal sign .Can someone please help me?
• Apr 30th 2011, 04:57 PM
topsquark
Quote:

Originally Posted by AkilMAI
How can I solve this equation $\displaystyle x''-x=(1+e^{t})^{-1}$ using the method of variation of parameters?Using the auxiliary equation the roots are $\displaystyle r_{1}=0$ and$\displaystyle r_{2}=1$.
I don't know if I'm suppose to look for a solution of the form $\displaystyle x(t)=p+q*e^{x}$ manly because the initial equation has o 1+... after the equal sign .Can someone please help me?

I'm not clear on how well you know the method, so I'll just say this for now. Recheck your solution to the homogeneous equation. m^2 - 1 = 0 gives what as solutions?

-Dan
• Apr 30th 2011, 05:31 PM
AkilMAI
m1=1 and m2=-1....I don't know why I had the impression that the equation is x''-x'=... and not x''-x=...
I know that this method is called variation of parameters manly because we have varied the usual parameters
,A and B for instance, and made them functions x(t)=u1(t)x1(t)+u2(t)x2(t)
• May 1st 2011, 12:20 PM
AkilMAI
but still I'm having trouble picking a solution to solve this
• May 1st 2011, 01:45 PM
topsquark
Quote:

Originally Posted by AkilMAI
m1=1 and m2=-1....I don't know why I had the impression that the equation is x''-x'=... and not x''-x=...
I know that this method is called variation of parameters manly because we have varied the usual parameters
,A and B for instance, and made them functions x(t)=u1(t)x1(t)+u2(t)x2(t)

Quote:

Originally Posted by AkilMAI
but still I'm having trouble picking a solution to solve this

Since m = 1 and m = - 1 you know that the homogeneous solutions are
$\displaystyle x_1(t) = e^t~\text{ and }~x_2(t) = e^{-t}$

So your proposed particular solution will be
$\displaystyle x_p(t) = u_1(t)e^t + u_2(t)e^{-t}$

Can you work from there?

-Dan