Question

linear D.E of nth order with constant coefficient...help guys
.

Answer 1

This WILL be messy... are you up for it? :D

Answer 2

yes

Answer 3

For these kinds of differential equations, the general form of the solution is given by
\[\Large f^{(h)}(x) + f^{(p)}(x)\] where the f^h is the general solution to the homogeneous equation while f^p is any particular solution of this equation. Can you find the general solution to the homogeneous equation?

Answer 4

\[m ^{^{2}}=11\]

Answer 5

What does that mean? D:

Answer 6

\[y=e ^{a}\]

Answer 7

Oh wait... there's no y' term?

Answer 8

get the higher order first D.E

Answer 9

So basically, it's
\[\Large y'' - 11y = xe^x\]?

Answer 10

\[m ^{2} +1 -12 =0\]

Answer 11

then m= square root of 11

Answer 12

then get the vlue of y

Answer 13

\[y=C{1}^{+\sqrt{11}x}+C{2}^{-\sqrt{11}x}\]

Answer 14

Im right? what is the next get the other y which from Xe^x

Answer 15

whic is y= Ae^u

Answer 16

take derivative until second derivative

Answer 17

ang get the value of A I stocked here

Answer 18

help guys if Im wrong :(

Answer 19

Well, if it doesn't work with only one undetermined coefficient, did you try 2?
Specifically this:
\[\Large y = \color{blue}Axe^x + \color{red}Be^x\]

Answer 20

subst the value of Y" ,Y' and Y to the y′′+ y -12y

Answer 21

where did you get this y=Axex+Bex

Answer 22

@UnkleRhaukus

Answer 23

are you sure the second term in the question isn't a derivative?

Answer 24

Odd, isn't it :D

Answer 25

Oh by the way, I didn't actually "get" that Axe^x + Be^x, for all intents and purposes, it IS just a guess... see if it gets you the coefficients you want, since Axe^x alone doesn't seem to do it.

Answer 26

if not (yes strange),
then i think you (@melmel ) got
the complementary homogenous solution
\[\large f^h(x)=y_c(x)=C_{1}e^{+\sqrt{11}x}+C_{2}e^{-\sqrt{11}x}\]
{forgot to type the e's in the bases }

Answer 27

Which method to calculate the particular solution
\[f^h(x)=y_p(x)\]
are you going to use; undetermined coefficients, operator D , or variation of parameters, @melmel ?

Answer 28

apparently offline :(
Oh well, onwards and upwards XD

Answer 29

^_^
thanks