Question

\[y^{\prime\prime}+4y^\prime+4y=16xe^{-2x}\]

Answer 1

\[y_c^{\prime\prime}+4y_c^\prime+4y_c=0\]
\[m^2+4m+4=0\]\[(m+2)^2=0\]\[m=-2\]
\[y_c=(A+Bx)e^{-2x}\]

Answer 2

\[y_p^{\prime\prime}+4y_p^\prime+4y_p=16xe^{-2x}\]
\[y_p=Cx^3e^{-2x}\]\[y_p^\prime= C(3x^2-2x^3)e^{-2x}\]\[y_p^{\prime\prime}=C(6x-6x^2-2(3x^2-2x^3))e^{-2x}=C(6x-12x^2+4x^3)e^{-2x}\]
\[C(6x-12x^2+4x^3)e^{-2x}+4(C(3x^2-2x^3)e^{-2x})+4(Cx^3e^{-2x})=16e^{-2x}\]\[C(6x-12x^2+4x^3+12x^2-8x^3+4x^3)e^{-2x}=16e^{-2x}\]\[6Cxe^{-2x}=16e^{-2x}\]\[C=\frac 83\]
\[y_p=\frac38x^3e^{-2x}\]

Answer 3

nice @UnkleRhaukus

Answer 4

no no why is your 8/3 suddenly turned to 3/8 lol

Answer 5

ok so have got the right answer,
but i dont understand why the trial for the particular
solution was assumed to be of this form\[y_p=Cx^3e^{-2x}\]

Answer 6

oh yeah i did not mean to invert the fraction for no reason

Answer 7

\[y_p=\frac83x^3e^{-2x}\]\[y=y_c+y_p\]\[y(x)=(A+Bx)e^{-2x}+\frac83x^3e^{-2x}\]\[y(x)=(A+Bx+\frac83x^3)e^{-2x}\]

Answer 8

how come \[y_p=Cx^2e^{-2x}\]
didn't work?

Answer 9

\[16xe^{-2x}\]
based on the table of undetermined coef
\[y_p = (Ax+B)e^{-2x}\]

if we time only x
\[Bxe^{-2x}\] will be the same with one of the homogenous solution

so we time x^2
\[y_p = x^2(Ax+B)e^{-2x}\]

Answer 10

wh not this?\[y_p=x(C+Dx)e^{-2x}\]

Answer 11

Cx e^(-2x) will be the same with the homogenous solution Bx e^(-2x)

Answer 12

dosent the Dx^2e^{-2x} have an extra factor of x though/?

Answer 13

^ doesn't matter tho o-o.., as long as it's not the same with the homogenous solution

Ae^(-2x) + Bx e^(-2x)

Answer 14

i dont get it

Answer 15

yes

Answer 16

is there some way we can know \[y_p = x^2(C+Dx)e^{-2x}\]
the C will be zero?

Answer 17

i don't know about that ._.

Answer 18

when there is two constants the differentiation becomes very confusing

Answer 19

so i should really go back and use\[y_p = x^2(C+Dx)e^{-2x}\]
and find \(y^\prime_p,\quad y_p^{\prime\prime}\) again and do the substitution again...

Answer 20

i check, your answer is correct tho o-o.. C probably will get cancelled out. but i don't know how to tell if C or D will get cancelled, tho.

Answer 21

\[y_p=(C+Dx)x^2e^{-2x}\]
\[y_p^\prime= (Dx^2+2(C+Dx)x-2(C+Dx)x^2)e^{-2x}\]\[\qquad= (2Cx-2Cx^2+3Dx^3-2Dx^3)e^{-2x}\]\[\qquad\qquad= \left(2C(x-x^2)+D(3x^3-2x^3)\right)e^{-2x}\]

Answer 22

i got the same so far :P

Answer 23

me too

Answer 24

*\[\qquad\qquad= \left(2C(x-x^2)+D(3x^2-2x^3)\right)e^{-2x}\]

Answer 25

ok.. the small 2 and 3 are very hard to different -_- sorry

Answer 26

here for you to check\[y_p''=(2C-8Cx+4Cx^2+6Dx-12Dx^2+4Dx^3)e^{-2x}\]

Answer 27

\[y_p^{\prime\prime}=\left(2C(1-2x)+D(6x-6x^2)-2\left(2C(x-x^2)+D(3x^2-2x^3)\right)\right)e^{-2x}\]\[=\left(2C-4Cx+6Dx-6Dx^2-4Cx+4Cx^2+6Dx^2-4Dx^3\right)e^{-2x}\]\[=\left(2C-8Cx+4Cx^2+6Dx-12Dx^2-4Dx^3\right)e^{-2x}\]

Answer 28

oh goodie i have what you have..
now i have substitute and solve for C,D

Answer 29

\[y_p^{\prime\prime}+4y_p^\prime+4y_p=16xe^{-2x}\]
\[y_p=(C+Dx)x^2e^{-2x}\]\[y_p^{\prime}=\left(2C(x-x^2)+D(3x^2-2x^3)\right)e^{-2x}\]\[y_p^{\prime\prime}=\left(C(2-8x+4x^2)+D(6x-12x^2-4x^3)\right)e^{-2x}\]

\[\tiny\left(C(2-8x+4x^2)+D(6x-12x^2-4x^3)\right)e^{-2x}+4\left(2C(x-x^2)+D(3x^2-2x^3)\right)e^{-2x}+4(C+Dx)x^2e^{-2x}=16xe^{-2x}\]

Answer 30

lol that ultra small text @_@

\[2Ce^{-2x}+6Dxe^{-2x} = 16xe^{-2x}\]

Answer 31

how did you simplify that so quick?

Answer 32

i.. wrote it down on paper, way faster than doing it in head or on comp lol

Answer 33

that has got to be an easier way,
there should be some way to know one of the constants is going to be zero somehow,,

Answer 34

\[y_p^{\prime\prime}+4y_p^\prime+4y_p=16xe^{-2x}\]
\[y_p=(C+Dx)x^2e^{-2x}\]
\[y_p^\prime= (Dx^2+2(C+Dx)x-2(C+Dx)x^2)e^{-2x}\]\[\qquad= (2Cx-2Cx^2+3Dx^2-2Dx^3)e^{-2x}\]\[\qquad\qquad= \left(2C(x-x^2)+D(3x^2-2x^3)\right)e^{-2x}\]
\[y_p^{\prime\prime}=\left(2C(1-2x)+D(6x-6x^2)-2\left(2C(x-x^2)+D(3x^2-2x^3)\right)\right)e^{-2x}\]\[=\left(2C-4Cx+6Dx-6Dx^2-4Cx+4Cx^2+6Dx^2-4Dx^3\right)e^{-2x}\]\[=\left(2C-8Cx+4Cx^2+6Dx-12Dx^2-4Dx^3\right)e^{-2x}\]\[=\left(C(2-8x+4x^2)+D(6x-12x^2-4x^3)\right)e^{-2x}\]
\[\begin{array}{ccc}
\left(C(2-8x+4x^2)+D(6x-12x^2-4x^3)\right)e^{-2x} \\
+4\left(2C(x-x^2)+D(3x^2-2x^3)\right)e^{-2x} \\
+4(C+Dx)x^2e^{-2x}
\end{array}=16xe^{-2x} \]
\[\left(\begin{array}{ccc}
C\left[(2-8x+4x^2)+8(x-x^2)+4x^2\right]\\
+D\left[(6x-12x^2-4x^3)+4\left(3x^2-2x^3\right)+4x^3\right]
\end{array}\right) e^{-2x}=16xe^{-2x}\]
\[\begin{array}{ccc}C\left[2-8x+4x^2+8x-8x^2+4x^2\right]\\+D\left[6x-12x^2-4x^3+12x^2-8x^3+4x^3\right]\end{array}=16x\]
\[2C+6Dx=16x\]
\[C=0;\qquad D=\frac 83\]
\[y_p=\frac83x^3e^{-2x}\]

Answer 35

\[y=y_c+y_p\]
\[y(x)=(A+Bx)e^{-2x}+\frac83x^3e^{-2x}\]\[y(x)=(A+Bx+\frac83x^3)e^{-2x}\]

Answer 36

surely there must be a more efficient way at finding the particular solution ,