anonymous
  • anonymous
find solution to following differential equation with initial conditions y''+y'-2y=6e^x-2x y(0)=2, y'(0)=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i am trying, give me some time, these problems are not short
anonymous
  • anonymous
do you have the answer from the text book?
anonymous
  • 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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anonymous
  • anonymous
then find \[y_p(x)=u_1(x)y_1(x) + u_2(x)y_2(x)\]
anonymous
  • anonymous
then you would need wronskian which i am still working on, lol...
anonymous
  • anonymous
\[w(y_1,y_2)=3e ^{-x}\] if anyone wants to correct any mistakes along the processes please feel free.. thanks
anonymous
  • anonymous
\[u_1=-(7/6)e ^{2x}-(x/3)e ^{2x}\] this problem is really long, i am about to give up. =(
anonymous
  • anonymous
i got y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.
anonymous
  • anonymous
start getting yc then the yp then y= yc+yp
anonymous
  • anonymous
\[u_2 = 2x - (2x/9)e ^{3x} - (2/27)e ^{3x}\]
anonymous
  • anonymous
thats why we ned to get the answer from the book itself...there are probably many solutions on this..
anonymous
  • 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
  • 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
  • anonymous
start w/ yc=c1e^x + c2e^-2x yp=Ae^x +Bxe^x+Cx
anonymous
  • anonymous
ok then there must be other methods that will make this shorter perhaps laplace? or annihilator?
anonymous
  • anonymous
i am done, i am happy i got that far til the y_p... haha.... sorry didn't help much!
anonymous
  • 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
  • anonymous
my method will still work, it just going to be long and most likely I already made a mistake
anonymous
  • anonymous
yes you can do it from LAPLACE and get the same answer
anonymous
  • anonymous
i could see lots lots lots of algebra involved with laplace!
anonymous
  • anonymous
after subing back in, i get \[3Be^x -2Cx+C=6e^x-2x\] any ideas ? haha
anonymous
  • anonymous
yes it might take sometimes in computing them
anonymous
  • anonymous
3B=6 -2C=-2 but C=0?
anonymous
  • anonymous
i got 2e^x +Be^x+C-2Cx=6e^x-2x
anonymous
  • anonymous
thats why ive been stuck on it, unless the intial conditions allow c=1
anonymous
  • anonymous
yes C=1
anonymous
  • anonymous
so whats A equal to? 0?
anonymous
  • anonymous
2e^x +Be^x+C-2Cx=6e^x-2x (2+B)e^x +C-2xC=6e^x-2x
anonymous
  • anonymous
yes A=0
anonymous
  • anonymous
did you get it nor rzawad?
anonymous
  • 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
  • anonymous
Yp=4xe^x +x Yc=C1e^x +C2e^-2x Y=Yc+Yp
anonymous
  • anonymous
gives Y=4xe^x +x+C1e^x+C2e^-2x
anonymous
  • anonymous
now you can take and subs the initial condition of the equation y(0)=2, y'(0)=0
anonymous
  • anonymous
so \[c_1+c_2 = 2?\]
anonymous
  • 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
  • anonymous
C1=2-C2=2-7/3=-1/3 now subs this to your eq
anonymous
  • 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
  • anonymous
end......
anonymous
  • anonymous
hope rzawad8 got the same solutions and answer..... lol
anonymous
  • anonymous
rzawad. where did you get this problem from?
anonymous
  • 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
  • anonymous
we worked hour+ for this problem and then look! Mathematica solved the D.E. in minute+ lol...
anonymous
  • anonymous
wow thats real cool whats the answer from matematica did you ty to solved it from there?
anonymous
  • anonymous
Even with a senior using Mathtematica it was more like and hour.
anonymous
  • anonymous
an hour.
anonymous
  • 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
  • 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
  • anonymous
if anyone here is taking DE now you can email me and i might help you out with your HW etc
anonymous
  • 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
  • anonymous
wow i didnt knowthere is wolfram ....lol
anonymous
  • anonymous
its from a past exam with no solutions dude soz,
anonymous
  • anonymous
ok hope you got an idea of solving of this type of prob...thnx

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