find solution to following differential equation with initial conditions
y''+y'-2y=6e^x-2x
y(0)=2, y'(0)=0

- anonymous

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- schrodinger

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- anonymous

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

- anonymous

do you have the answer from the text book?

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

start w/ yc=c1e^x + c2e^-2x
yp=Ae^x +Bxe^x+Cx

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes it might take sometimes in computing them

- anonymous

3B=6
-2C=-2
but C=0?

- anonymous

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

- anonymous

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

- anonymous

yes C=1

- anonymous

so whats A equal to? 0?

- anonymous

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

- anonymous

yes A=0

- anonymous

did you get it nor rzawad?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

so \[c_1+c_2 = 2?\]

- anonymous

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

- anonymous

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

- anonymous

Y=4xe^x +x+C1e^x+C2e^-2x subst here gives
y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.

- anonymous

end......

- anonymous

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

- anonymous

rzawad. where did you get this problem from?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

an hour.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

wow i didnt knowthere is wolfram ....lol

- anonymous

its from a past exam with no solutions dude soz,

- anonymous

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

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