Any Maths Genius Out There- I Need Ur Help!!!!!!

[vannistelrooy]

please can someone help me?

[vb2dub]

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

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

[Quantumcat]

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 :-)

[Quantumcat]

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!