Question

\[y'' - 4 y' +4y = e^{2x}\]

Answer 1

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

Answer 2

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

Answer 3

\[\text{The homogenous solution has an }e^{2x} \text{ and an }xe^{2x} \text{ term already};\]\[\text{ the particular solution must be higher order}\]

\[y_p=Cx^2e^{2x}\]\[ y_p^\prime=2Cxe^{2x}+2Cx^2e^{2x}=2Cx(1+x)e^{2x}\]\[ y_p^{\prime\prime}=2Ce^{2x}+4Cxe^{2x}+4Cxe^{2x}+4Cx^2e^{2x}=2C(2x^2+4x+1)e^{2x}\]
right so far?

Answer 4

have i chosen the right form of for the particular solution?

Answer 5

maybe i should have two constants?
\[y_p=(C+Dx)x^2e^{2x}\]?

Answer 6

\(y_{p} = Cx^{2}e^{2x}\) is correct.

Answer 7

\[\left(2C(2x^2+4x+1)e^{2x}\right)-4\left(2Cx(1+x)e^{2x}\right)+4(Cx^2e^{2x})=e^{2x}\]\[2C\left((2x^2+4x+1)-4x(1+x)+2x^2\right)e^{2x}=e^{2x}\]\[2C\left(2x^2+4x+1-4x-4x^2+2x^2\right)e^{2x}=e^{2x}\]\[2Ce^{2x}=e^{2x}\]\[C=\frac 12\]

Answer 8

\[y_p=\frac {x^2}2e^{2x}\]

Answer 9

\[y(x)=(A+Bx)e^{2x}+\frac{x^2}2e^{2x}\]?

Answer 10

yes.

Answer 11

oh

Answer 12

are you sure
,

Answer 13

http://www.wolframalpha.com/input/?i=y%27%27+-4y%27+%2B+4y+%3D+e%5E%282x%29

Answer 14

ah, i had typed the rong thing into wolfram when i tried , thanks

Answer 15

http://www.wolframalpha.com/input/?i=y%27%27-4y%27%2B4%3De%5E2x
lolz