Question

\[y_p^{\prime\prime}+4y_p^\prime+4y_p=12x^2e^{-2x}\]

Answer 1

\[y_p=(C+Dx)xe^{-2x}\]
\[y^\prime_p=(Dx+C+Dx-2x(C+Dx))e^{-2x}\]\[\qquad\quad =(Dx+C+Dx-2Cx-2Dx^2)e^{-2x}\]\[\qquad\quad =(C+2Dx-2Cx-2Dx^2)e^{-2x}\]
\[y^{\prime\prime}_p=\left(2D-2C-4Dx-2(C+2Dx-2Cx-2Dx^2)\right)e^{-2x}\]\[\qquad\qquad\quad=\left(2D-2C-4Dx-2Dx-2C-2Dx+4Cx+4Dx^2\right)e^{-2x}\]\[\qquad\qquad\quad=\left(2D-4C-8Dx+4Cx+4Dx^2\right)e^{-2x}\]
\[\begin{array}{ccc}\left(2D-4C-8Dx+4Cx+4Dx^2\right)e^{-2x}\\+4(C+2Dx-2Cx-2Dx^2)e^{-2x}\\+4(C+Dx)xe^{-2x}\end{array}=12x^2e^{-2x}\]
\[\left(\begin{array}{ccc} 2D-4C-8Dx+4Cx+4Dx^2 \\+4C+8Dx-8Cx-8Dx^2\\ \qquad\qquad+4Cx+4Dx^2\end{array}\right)e^{-2x}=12x^2e^{-2x}\]
\[2D=12x^2\]

Answer 2

something went wrong

Answer 3

\[y_p \text{ should be } y_p=Cx^2e^{-2x}+Dx^3e^{-2x}+Ex^4e^{-2x}\]since e^-2x and xe^(-2x) is in the homogeneous solution

Answer 4

agh

Answer 5

?

Answer 6

\[y_p=Cx^2e^{-2x}+Dx^3e^{-2x}+Ex^4e^{-2x}\]
\[y_p'=2Cxe^{-2x}-2Cx^2e^{-2x}+3Dx^2e^{-2x}-2Dx^3e^{-2x}+4Ex^3e^{-2x}-2Ex^4e^{-2x}\]
\[y_p''=2Ce^{-2x}-4Cxe^{-2x}-4Cxe^{-2x}+4Cx^2e^{-2x}+6Dxe^{-2x}-6Dx^2e^{-2x}-6Dx^2e^{-2x}\]\[+4Dx^3e^{-2x}+12Ex^2e^{-2x}-8Ex^3e^{-2x}-8Ex^3e^{-2x}+4Ex^4e^{-2x}=\]
\[2Ce^{-2x}-8Cxe^{-2x}+4Cx^2e^{-2x}+6Dxe^{-2x}-12Dx^2e^{-2x}+4Dx^3e^{-2x}\]\[+12Ex^2e^{-2x}-16Ex^3e^{-2x}+4Ex^4e^{-2x}\]

Answer 7

isn't this a better particular solution?
\[y_p=(Cx+Dx^2+Ex^3)e^{-2x}\]

Answer 8

i guess not, dammnit

Answer 9

\[y_p=(Cx^2+Dx^3+Ex^4)e^{-2x}\]
\[y^\prime_p=\left(2Cx+3Dx^2+4Ex^3-2(Cx^2+Dx^3+Ex^4)\right)e^{-2x}\]\[\qquad\quad =\left(2Cx+3Dx^2+4Ex^3-2Cx^2-2Dx^3-2Ex^4\right)e^{-2x}\]

Answer 10

\[\Tiny y^{\prime\prime}_p=\left(2C+6Dx+12Ex^2-2Cx-6Dx^2-8Ex^3-2(2Cx+3Dx^2+4Ex^3-2Cx^2-2Dx^3-2Ex^4)\right)e^{-2x}\]\[\Tiny \qquad=\left(2C+6Dx+12Ex^2-2Cx-6Dx^2-8Ex^3-4Cx-6Dx^2-8Ex^3+4Cx^2+4Dx^3+4Ex^4\right)e^{-2x}\]\[\Tiny \qquad=\left(2C+6Dx+12Ex^2-6Cx-12Dx^2-16Ex^3+4Cx^2+4Dx^3+4Ex^4\right)e^{-2x}\]

Answer 11

our derivatives are in agreement

Answer 12

yours has -6Cx while mine is -8Cx
*checking*

Answer 13

oh yeah

Answer 14

-8Cx, will cancel better,

Answer 15

i see my mistake , it on the first line of the second derivative
the fourth term
-2Cx should be-4Cx

Answer 16

\[\tiny y^{\prime\prime}_p=\left(2C+6Dx+12Ex^2-4Cx-6Dx^2-8Ex^3-2(2Cx+3Dx^2+4Ex^3-2Cx^2-2Dx^3-2Ex^4)\right)e^{-2x}\]\[\tiny\qquad=\left(2C+6Dx+12Ex^2-4Cx-6Dx^2-8Ex^3-4Cx-6Dx^2-8Ex^3+4Cx^2+4Dx^3+4Ex^4\right)e^{-2x}\]\[\tiny\qquad=\left(2C+6Dx+12Ex^2-8Cx-12Dx^2-16Ex^3+4Cx^2+4Dx^3+4Ex^4\right)e^{-2x}\]

Answer 17

yea

Answer 18

\[\begin{array}{ccc}\left(2C+6Dx+12Ex^2-8Cx-12Dx^2-16Ex^3+4Cx^2+4Dx^3+4Ex^4\right)e^{-2x}
\\+4\left(2Cx+3Dx^2+4Ex^3-2Cx^2-2Dx^3-2Ex^4\right)e^{-2x}
\\+4(Cx^2+Dx^3+Ex^4)e^{-2x}\end{array}=12x^2e^{-2x}\]
\[\left(\begin{array}{ccc} 2C+6Dx+12Ex^2-8Cx-12Dx^2-16Ex^3+4Cx^2+4Dx^3+4Ex^4
\\ \qquad\qquad\qquad\qquad+8Cx+12Dx^2+16Ex^3-8Cx^2-8Dx^3-8Ex^4
\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+4Cx^2+4Dx^3+4Ex^4\end{array}\right)e^{-2x}=12x^2e^{-2x}\]

\[(2C+6Dx+12Ex^2)=12x^2\]

Answer 19

C=0
D=0
E=1

Answer 20

\[y_p=x^4e^{-2x}\]

Answer 21

\[y(x)=(A+Bx)e^{-2x}+x^4e^{-2x}\qquad\checkmark\]

Answer 22

thanks for your help @nphuongsun93

Answer 23

ty ty