Question

Solve by undermined coefficients/variation of parameters:
\[ y''-4y'+4y = \frac{e^{2x}}{x}\]

Answer 1

\[ y_c''-4y_c'+4y_c=0\]\[ \lambda^2-4\lambda+4=0\]\[ \lambda=2, 2\]\[y_c = c_1e^{2x}+c_2xe^{2x}\]
How can I guess the particular solution?

Answer 2

you can't do this one with undetermined cofficients. you would have to use variation of parameter.

Answer 3

to find the yp

Answer 4

Why doesn't undetermined coefficients work in this case??

Answer 5

well in undetermined cofficients there are few general forms of the solution already told.. i mean like if f(x) is
e^x = Ae^ax.. and a few more like these.. know of it?

Answer 6

for "x" general form of its solution of yp would be Ax+B

Answer 7

Is xe^(2x) a general form?

Answer 8

general for have unknown cofficients.. this one is not general form.. but this has a general for , it would be (Ax+B)e^(cx)

Answer 9

Then how do you define ''general''??

Answer 10

like for equation of a line y=mx+c.. this is general form. u ahve to find x and m.. same way in here yp = (Ax+B)e^(cx) is general form and we have to find A,B,c. th

Answer 11

Some typical examples are
constant - constant
sin / cos - Asinx + Bcosx
e^(ax) = Ce^(ax)
x^n = polynomial of x

But xe^(2x) is not that ''general''

Answer 12

in exams we never get the straight one .. in here we have to first see if x has general for or not.. then we would see if e^2x has general form or nto.. if they both would have just miltiply them

Answer 13

\(x e^{2x}\) still qualifies as general

Answer 14

and e^(2x) / x on the other hand is not general?

Answer 15

in my opinion, No. because i think that x in the denominator part is the trouble here and we dont have a general form for that part. so i think variation of parameter should do the trick.

Answer 16

Variation of parameters:

\[y'' - 4y' + 4y = \frac{e^{2x}}{x}\]\[y_c'' - 4y_c' + 4y_c =0\]\[y_c = c_1e^{2x}+c_2xe^{2x}\]
\[u_1'e^{2x} + u_2' xe^{2x}=0\]\[2u_1'e^{2x} + u_2' (2xe^{2x}+e^{2x})=\frac{e^{2x}}{x}\]
\[u_2'e^{2x}=\frac{e^{2x}}{x}\]\[u_2 = lnx\]
\[u_1' = -\frac{\frac{1}{x}xe^{2x}}{e^2x}\]\[u_1 = -x\]
\[y=c_1e^{2x}+c_2xe^{2x} +xlnxe^{2x} - xe^{2x}\]
I can't believe that I got the solution! :O

Answer 17

lol well i am glad it worked out for u :)

Answer 18

lol Thanks!! :)