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Mathematics 68 Online
OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

that sounds like a lot of work

OpenStudy (anonymous):

lol it is

OpenStudy (unklerhaukus):

bother, and thankyou

OpenStudy (unklerhaukus):

do i really need Three constants ?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

that will save me some work

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

What if you take the particular solution as: \[y_p = Ce^{-x} + De^{-3x} + Exe^x\] Try this if it helps..

OpenStudy (anonymous):

somehow the y' looks wrong @UnkleRhaukus @waterineyes there is also x e^x in the homogeous solution, think it won't work

OpenStudy (unklerhaukus):

y'_p looks right to me

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

that sound easier @eliassaab

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

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

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

i know the solution im just having trouble with the steps

OpenStudy (unklerhaukus):

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

OpenStudy (unklerhaukus):

\[y=y_c+y_{p_1}+y_{p_2}+y_{p_3}\] \[y(x)=(A+Bx)e^x+\frac 32e^{-x}+\frac 14e^{-3x}+\frac{x^3}6e^x\] got it!

OpenStudy (anonymous):

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

OpenStudy (unklerhaukus):

your missing an x @nphuongsun93

OpenStudy (unklerhaukus):

your F is my E

OpenStudy (anonymous):

oh ya i seeeeeeeeeeeeeee it

OpenStudy (unklerhaukus):

i have leant it much easier with one constant

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