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anonymous
 5 years ago
find solution to following differential equation with initial conditions
y''+y'2y=6e^x2x
y(0)=2, y'(0)=0
anonymous
 5 years ago
find solution to following differential equation with initial conditions y''+y'2y=6e^x2x y(0)=2, y'(0)=0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am trying, give me some time, these problems are not short

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have the answer from the text book?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01st solve y"+y'2y=0 Charateristic Eq. m^2 + m  2 = 0 (m+2)(m1)=0 m=2, m=1 so \[y_c = c_1 e ^{2x} + c_2 e ^{?}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then find \[y_p(x)=u_1(x)y_1(x) + u_2(x)y_2(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you would need wronskian which i am still working on, lol...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[w(y_1,y_2)=3e ^{x}\] if anyone wants to correct any mistakes along the processes please feel free.. thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[u_1=(7/6)e ^{2x}(x/3)e ^{2x}\] this problem is really long, i am about to give up. =(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got y=4xe^x +x (e^x)/3 +(7e^2x)/3 G.S.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0start getting yc then the yp then y= yc+yp

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[u_2 = 2x  (2x/9)e ^{3x}  (2/27)e ^{3x}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats why we ned to get the answer from the book itself...there are probably many solutions on this..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[y_p = [(7/6)e ^{2x}  (x/3)e ^{2x} ][e ^{2x}]+[2x(2x/9)e ^{3x}(2/27)e ^{3x}][e ^{x}]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0start w/ yc=c1e^x + c2e^2x yp=Ae^x +Bxe^x+Cx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok then there must be other methods that will make this shorter perhaps laplace? or annihilator?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am done, i am happy i got that far til the y_p... haha.... sorry didn't help much!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yp'=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^x2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my method will still work, it just going to be long and most likely I already made a mistake

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes you can do it from LAPLACE and get the same answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i could see lots lots lots of algebra involved with laplace!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after subing back in, i get \[3Be^x 2Cx+C=6e^x2x\] any ideas ? haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it might take sometimes in computing them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got 2e^x +Be^x+C2Cx=6e^x2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats why ive been stuck on it, unless the intial conditions allow c=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so whats A equal to? 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02e^x +Be^x+C2Cx=6e^x2x (2+B)e^x +C2xC=6e^x2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you get it nor rzawad?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02e^x +Be^x+C2Cx=6e^x2x (2+B)e^x +C2xC=6e^x2x 2+B=6 > B=4 and 2xC=2x >C=1 now subs to Yp

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yp=4xe^x +x Yc=C1e^x +C2e^2x Y=Yc+Yp

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gives Y=4xe^x +x+C1e^x+C2e^2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you can take and subs the initial condition of the equation y(0)=2, y'(0)=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Y(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+C12C2 0=5+C12C2 sub C1=2C2 0=5+2C22C2 C2=7/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0C1=2C2=27/3=1/3 now subs this to your eq

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Y=4xe^x +x+C1e^x+C2e^2x subst here gives y=4xe^x +x (e^x)/3 +(7e^2x)/3 G.S.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hope rzawad8 got the same solutions and answer..... lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0rzawad. where did you get this problem from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y[x]=\frac{1}{2}+\frac{3 e^{2 x}}{2}+x+2 e^x x \] \[y\text{''}[x]+y'[x]2y[x]\text{=}6E{}^{\wedge}x2x\text{//}\text{Simplify} \] \[\left\{6 e^x2 x\right\}=6 e^x2 x \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we worked hour+ for this problem and then look! Mathematica solved the D.E. in minute+ lol...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow thats real cool whats the answer from matematica did you ty to solved it from there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Even with a senior using Mathtematica it was more like and hour.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i just solved this prob in 15 mins lol..and hoping to get the answer from the text book from rzawad..lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0last 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
 5 years ago
Best ResponseYou've already chosen the best response.0if anyone here is taking DE now you can email me and i might help you out with your HW etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=y%27%27%2By%272y%3D6e%5Ex2x+%2C+y%280%29%3D2+%2C+y%27%280%29%3D0 wolfram will give the same answer as robtobey's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow i didnt knowthere is wolfram ....lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its from a past exam with no solutions dude soz,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok hope you got an idea of solving of this type of prob...thnx
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