find solution to following differential equation with initial conditions y''+y'-2y=6e^x-2x y(0)=2, y'(0)=0
i am trying, give me some time, these problems are not short
do you have the answer from the text book?
1st solve y"+y'-2y=0 Charateristic Eq. m^2 + m - 2 = 0 (m+2)(m-1)=0 m=-2, m=1 so \[y_c = c_1 e ^{-2x} + c_2 e ^{?}\]
then find \[y_p(x)=u_1(x)y_1(x) + u_2(x)y_2(x)\]
then you would need wronskian which i am still working on, lol...
\[w(y_1,y_2)=3e ^{-x}\] if anyone wants to correct any mistakes along the processes please feel free.. thanks
\[u_1=-(7/6)e ^{2x}-(x/3)e ^{2x}\] this problem is really long, i am about to give up. =(
i got y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.
start getting yc then the yp then y= yc+yp
\[u_2 = 2x - (2x/9)e ^{3x} - (2/27)e ^{3x}\]
thats why we ned to get the answer from the book itself...there are probably many solutions on this..
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?
\[y_p = [-(7/6)e ^{2x} - (x/3)e ^{2x} ][e ^{-2x}]+[2x-(2x/9)e ^{3x}-(2/27)e ^{3x}][e ^{x}]\]
start w/ yc=c1e^x + c2e^-2x yp=Ae^x +Bxe^x+Cx
ok then there must be other methods that will make this shorter perhaps laplace? or annihilator?
i am done, i am happy i got that far til the y_p... haha.... sorry didn't help much!
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
my method will still work, it just going to be long and most likely I already made a mistake
yes you can do it from LAPLACE and get the same answer
i could see lots lots lots of algebra involved with laplace!
after subing back in, i get \[3Be^x -2Cx+C=6e^x-2x\] any ideas ? haha
yes it might take sometimes in computing them
3B=6 -2C=-2 but C=0?
i got 2e^x +Be^x+C-2Cx=6e^x-2x
thats why ive been stuck on it, unless the intial conditions allow c=1
yes C=1
so whats A equal to? 0?
2e^x +Be^x+C-2Cx=6e^x-2x (2+B)e^x +C-2xC=6e^x-2x
yes A=0
did you get it nor rzawad?
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
Yp=4xe^x +x Yc=C1e^x +C2e^-2x Y=Yc+Yp
gives Y=4xe^x +x+C1e^x+C2e^-2x
now you can take and subs the initial condition of the equation y(0)=2, y'(0)=0
so \[c_1+c_2 = 2?\]
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
C1=2-C2=2-7/3=-1/3 now subs this to your eq
Y=4xe^x +x+C1e^x+C2e^-2x subst here gives y=4xe^x +x -(e^x)/3 +(7e^-2x)/3 G.S.
end......
hope rzawad8 got the same solutions and answer..... lol
rzawad. where did you get this problem from?
\[y[x]=\frac{1}{2}+\frac{3 e^{-2 x}}{2}+x+2 e^x x \] \[y\text{''}[x]+y'[x]-2y[x]\text{=}6E{}^{\wedge}x-2x\text{//}\text{Simplify} \] \[\left\{6 e^x-2 x\right\}=6 e^x-2 x \]
we worked hour+ for this problem and then look! Mathematica solved the D.E. in minute+ lol...
wow thats real cool whats the answer from matematica did you ty to solved it from there?
Even with a senior using Mathtematica it was more like and hour.
an hour.
well i just solved this prob in 15 mins lol..and hoping to get the answer from the text book from rzawad..lol
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
if anyone here is taking DE now you can email me and i might help you out with your HW etc
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
wow i didnt knowthere is wolfram ....lol
its from a past exam with no solutions dude soz,
ok hope you got an idea of solving of this type of prob...thnx
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