please can someone help me?
http://i82.photobucket.com/albums/j2...n1000/4028.jpg
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please can someone help me?
http://i82.photobucket.com/albums/j2...n1000/4028.jpg
if the answer is 4
i never passed MATH A in highschool so I dont know math that well but I removed all the numerators and denominators junk..
and what I see is this.
2 + 0 = 4
and since t = 0 and 0 with line in middle is 0
IDK what d is
im amusing d = 2 lol..
now if i plug em in
Code:2square and 0 2 and 0
--------- + 2 ------------ + 0 = 4
2 and 0sqaure 2 and 0
4 0
--- + 2 --- + 0 = 4
0 0
4 2 0
--- + --- --- + 0 = 4
0 1 0
IDK how to get any deeper bro
Hi there, I am doing a course on DE's at the moment, maybe I can help!
It is nonhomogeneous, and the solution to a non-homogeneous is the homogeneous linear combination of solutions + the particular solution.
Homogeneous solution is found by taking the characteristic equation:
s^2 + 2s + 1 = 0 (I can't write lambda so I'll use s instead)
(s+1)^2 = 0
Has a double root at s=-1
So the homogeneous solution is Theta=c1 e^-t + c2te^-t
For the particular solution we want
yp'' + yp' + yp = 4
since the nonhomogeneous term is constant this is easy! Don't haave to worry about method by variation of parameters or undetermined coefficients (the only two I know)
just set yp = 4, cause yp' and yp'' are zero.
So your solution is,
Theta = c1 e^-t + c2te^-t + 4
Easy!
This is a very straightforward problem if you are doing an introductory course on differential equations, but if you aren't I can't see where you got the problem from. Anyway, it was useful doing the problem, helped me revise for my exam :-)
Oops I didn't noticce it was an IVP. Easy though.
Theta = c1 e^-t + c2te^-t + 4
t=0. theta = 0, theta'=0, t=0
0= c1*1 + c2*0*1 + 4
c1=-4
theta' = -c1 e^-t + -c2te^-t+c2e^-t + 0
0=-c1*1 + 0*c2*1 + c2*1 + 0
0=4+c2
c2=-4 too
So now the solution with the initial conditions, is
Theta = -4e^-t -4te^-t + 4
Done!