Question

find solution to following differential equation with initial conditions

y''+y'-2y=6e^x-2x

y(0)=2, y'(0)=0

Answer 1

i am trying, give me some time, these problems are not short

Answer 2

do you have the answer from the text book?

Answer 3

1st solve y"+y'-2y=0
Charateristic Eq. m^2 + m - 2 = 0
(m+2)(m-1)=0
m=-2, m=1
so \[y_c = c_1 e ^{-2x} + c_2 e ^{?}\]

Answer 4

then find \[y_p(x)=u_1(x)y_1(x) + u_2(x)y_2(x)\]

Answer 5

then you would need wronskian
which i am still working on, lol...

Answer 6

\[w(y_1,y_2)=3e ^{-x}\]
if anyone wants to correct any mistakes along the processes please feel free..
thanks

Answer 7

\[u_1=-(7/6)e ^{2x}-(x/3)e ^{2x}\]
this problem is really long, i am about to give up. =(

Answer 8

i got y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.

Answer 9

start getting yc then the yp then y= yc+yp

Answer 10

\[u_2 = 2x - (2x/9)e ^{3x} - (2/27)e ^{3x}\]

Answer 11

thats why we ned to get the answer from the book itself...there are probably many solutions on this..

Answer 12

its not really that long!, im pretty sure we can use the method of undetermined coeffeiients but i cant seem to get a right answer to fit, i think my yp is wrong, variation of paremeters might be unessecary?

Answer 13

\[y_p = [-(7/6)e ^{2x} - (x/3)e ^{2x} ][e ^{-2x}]+[2x-(2x/9)e ^{3x}-(2/27)e ^{3x}][e ^{x}]\]

Answer 14

start w/ yc=c1e^x + c2e^-2x

yp=Ae^x +Bxe^x+Cx

Answer 15

ok then there must be other methods that will make this shorter
perhaps laplace?
or annihilator?

Answer 16

i am done, i am happy i got that far til the y_p... haha.... sorry didn't help much!

Answer 17

yp'=Ae^x +B(xe^x+e^x)+C
yp"=Ae^x+B(xe^x+e^x+e^x) now sbstitute this to the orig equation
y''+y'-2y=6e^x-2x

Answer 18

my method will still work, it just going to be long and most likely I already made a mistake

Answer 19

yes you can do it from LAPLACE and get the same answer

Answer 20

i could see lots lots lots of algebra involved with laplace!

Answer 21

after subing back in, i get
\[3Be^x -2Cx+C=6e^x-2x\]
any ideas ? haha

Answer 22

yes it might take sometimes in computing them

Answer 23

3B=6
-2C=-2

but C=0?

Answer 24

i got 2e^x +Be^x+C-2Cx=6e^x-2x

Answer 25

thats why ive been stuck on it, unless the intial conditions allow c=1

Answer 26

yes C=1

Answer 27

so whats A equal to? 0?

Answer 28

2e^x +Be^x+C-2Cx=6e^x-2x

(2+B)e^x +C-2xC=6e^x-2x

Answer 29

yes A=0

Answer 30

did you get it nor rzawad?

Answer 31

2e^x +Be^x+C-2Cx=6e^x-2x
(2+B)e^x +C-2xC=6e^x-2x
2+B=6 -> B=4
and -2xC=-2x -->C=1 now subs to Yp

Answer 32

Yp=4xe^x +x
Yc=C1e^x +C2e^-2x
Y=Yc+Yp

Answer 33

gives Y=4xe^x +x+C1e^x+C2e^-2x

Answer 34

now you can take and subs the initial condition of the equation
y(0)=2, y'(0)=0

Answer 35

so \[c_1+c_2 = 2?\]

Answer 36

Y(0)=C1+C2=2 and take the Y'=4(xe^x +e^x) +1 +C1e^x+C2e^-2x
take Y'(o)=0=4(1)+1+C1-2C2
0=5+C1-2C2
sub C1=2-C2 0=5+2-C2-2C2
C2=7/3

Answer 37

C1=2-C2=2-7/3=-1/3 now subs this to your eq

Answer 38

Y=4xe^x +x+C1e^x+C2e^-2x subst here gives

y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.

Answer 39

end......

Answer 40

hope rzawad8 got the same solutions and answer..... lol

Answer 41

rzawad. where did you get this problem from?

Answer 42

\[y[x]=\frac{1}{2}+\frac{3 e^{-2 x}}{2}+x+2 e^x x \] \[y\text{''}[x]+y'[x]-2y[x]\text{=}6E{}^{\wedge}x-2x\text{//}\text{Simplify} \] \[\left\{6 e^x-2 x\right\}=6 e^x-2 x \]

Answer 43

we worked hour+ for this problem
and then look! Mathematica solved the D.E. in minute+
lol...

Answer 44

wow thats real cool whats the answer from matematica did you ty to solved it from there?

Answer 45

Even with a senior using Mathtematica it was more like and hour.

Answer 46

an hour.

Answer 47

well i just solved this prob in 15 mins lol..and hoping to get the answer from the text book from rzawad..lol

Answer 48

last sem i had this Advanced engineering math even though we had a DE subject last sem. we had to cover this type of problems

Answer 49

if anyone here is taking DE now you can email me and i might help you out with your HW etc

Answer 50

http://www.wolframalpha.com/input/?i=y%27%27%2By%27-2y%3D6e%5Ex-2x+%2C+y%280%29%3D2+%2C+y%27%280%29%3D0

wolfram will give the same answer as robtobey's

Answer 51

wow i didnt knowthere is wolfram ....lol

Answer 52

its from a past exam with no solutions dude soz,

Answer 53

ok hope you got an idea of solving of this type of prob...thnx