Question

\[y^{\prime\prime}-2y^\prime+y=6e^{-x}+4e^{-3x}+xe^x\]
\[\text{---------}\]

Answer 1

\[y_c^{\prime\prime}-2y_c^\prime+y_c=0\]
\[m^2-2m+1=0\]\[(m-1)^2=\]\[m=1\]
\[y_c=(A+Bx)e^x\]\[\text{---------}\]

Answer 2

\[y_{p_1}^{\prime\prime}-2y_{p_1}^\prime+y_{p_1}=6e^{-x}\]
\[y_{p_1}=Ce^{-x};\qquad y^{\prime}_{p_1}=-Ce^{-x};\qquad y^{\prime\prime}_{p_1}=Ce^{-x}\]
\[(Ce^{-x})-2(-Ce^{-x})+(Ce^{-x})=6e^{-x}\]\[C(1+2+1)e^{-x}=6e^{-x}\]\[C=\frac 32\]
\[y_{p_1}=\frac 32e^{-x}\]\[\text{---------}\]

Answer 3

\[y_{p_2}^{\prime\prime}-2y_{p_2}^\prime+y_{p_2}=4e^{-3x}\]
\[y_{p_2}=De^{-3x};\qquad y_{p_2}=-3De^{-3x};\qquad y_{p_3}=9De^{-3x}\]
\[(9De^{-3x})-2(-3De^{-3x})+(De^{-3x})=4e^{-3x}\]\[D(9+6+1)e^{-3x}=4e^{-3x}\]\[D=\frac 14\]
\[y_{p_2}=\frac 14e^{-3x}\]\[\text{---------}\]

Answer 4

\[y_{p_3}^{\prime\prime}-2y_{p_3}^\prime+y_{p_3}=xe^x\]
\[y_{p_3}=Ex^2e^{x};\qquad y^\prime_{p_3}=2Exe^{x}+Ex^2e^x;\qquad y_{p_3}^{\prime\prime}=2Ee^x+2Exe^x+2Exe^x+Ex^2e^x\]
\[(2Ee^x+2Exe^x+2Exe^x+Ex^2e^x)-2(2Exe^{x}+Ex^2e^x)+(Ex^2e^{x})=xe^x\]\[E(2+2x+2x+x^2-4x-2x^2+x^2)e^x=xe^x\]

\[2Ee^x=xe^x\]
?

Answer 5

i must have made a mistake in when trying to solve \(y_{p_3}\),
can you find the mistake?

Answer 6

(Ex^2 + Fx + G)e^x

Answer 7

that sounds like a lot of work

Answer 8

lol it is

Answer 9

bother, and thankyou

Answer 10

do i really need Three constants ?

Answer 11

i think so, since it's x^2. it will cancel out at the end.

Answer 12

wait actually it's only 2 constants
x^2 (Ex + F)e^x <----------------------------!!

Answer 13

that will save me some work

Answer 14

\[y_{p_3}^{\prime\prime}-2y_{p_3}^\prime+y_{p_3}=xe^x\]
\[y_{p_3}=(E+Fx)x^2e^{x}\]
\[y^\prime_{p_3}=(Fx^2+2(E+Fx)x+(E+Fx)x^2)e^x\]
\[\small y_{p_3}^{\prime\prime}=\left(2Fx+2(Fx+(E+Fx))+Fx^2+2(E+Fx)x))+(Fx^2+2(E+Fx)x+(E+Fx)x^2)\right)e^x\]\[=(6Fx+4E+6Fx^2+2Ex+Ex^2+Fx^3)e^x\]\[=\left(Fx^3+(E+6F)x^2+(6F+2E)x+4E\right)e^x\]

Answer 15

What if you take the particular solution as:

\[y_p = Ce^{-x} + De^{-3x} + Exe^x\]

Try this if it helps..

Answer 16

somehow the y' looks wrong @UnkleRhaukus

@waterineyes there is also x e^x in the homogeous solution, think it won't work

Answer 17

y'_p looks right to me

Answer 18

\[\huge Ex^2e^x+Fx^3e^x\]
\[\huge y_p' = 2Exe^x + Ex^2 e^x + 3Fx^2e^x + Fx^3e^x\]

Answer 19

Take the particular solution for x e^x to be C x^3 e^x and you fine C=1/6

Answer 20

that sound easier @eliassaab

Answer 21

i got 1/6 xe^x in the end :s

Answer 22

Your complete particular solution is
\[
\frac{e^x x^3}{6}+\frac{e^{-3
x}}{4}+\frac{3 e^{-x}}{2}
\]

Answer 23

yeah i have the answer in the back of the book , i know E=frac16

Answer 24

\[y_{p_3}=Ex^3e^x\]\[y^\prime_{p_3}=E(3x^2+x^3)e^x\]\[y^{\prime\prime}=E(6x+3x^2+3x^2+x^3)e^x\]

Answer 25

The solution is
\[
c_1 e^x+c_2 e^x x+\frac{e^x
x^3}{6}+\frac{e^{-3
x}}{4}+\frac{3 e^{-x}}{2}
\]

Answer 26

i know the solution im just having trouble with the steps

Answer 27

\[E(6x+3x^2+3x^2+x^3)e^x-2E(3x^2+x^3)e^x+Ex^3e^x=xe^x\]\[E\left(6x+6x^2+x^3-6x^2-2x^3+x^3\right)e^x=xe^x\]\[6Exe^x=xe^x\]\[E=\frac 16\]
\[y_{p_3}=\frac{x^3}6e^x\]

Answer 28

\[y=y_c+y_{p_1}+y_{p_2}+y_{p_3}\]

\[y(x)=(A+Bx)e^x+\frac 32e^{-x}+\frac 14e^{-3x}+\frac{x^3}6e^x\]

got it!

Answer 29

i'm sticking with my way tho!
\[y = Ex^2e^x + Fx^3e^x\]
\[y'=2Exe^x +Ex^2e^x + 3Fx^2e^x+Fx^3e^x\]
\[y'' = 2Ee^x+4Exe^x+Ex^2e^x+6Fxe^x+6x^2e^x+Fx^3e^x\]

y''-2y'+y=xe^x
\[2Ee^x+6Fxe^x = xe^x\]
\[F= \frac{1}{6}\]
\[y = \frac{x^2e^x}{6}\]

Answer 30

your missing an x @nphuongsun93

Answer 31

your F is my E

Answer 32

oh ya i seeeeeeeeeeeeeee it

Answer 33

i have leant it much easier with one constant