anonymous
  • anonymous
Solve by undermined coefficients/variation of parameters: \[ y''-4y'+4y = \frac{e^{2x}}{x}\]
Differential Equations
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[ y_c''-4y_c'+4y_c=0\]\[ \lambda^2-4\lambda+4=0\]\[ \lambda=2, 2\]\[y_c = c_1e^{2x}+c_2xe^{2x}\] How can I guess the particular solution?
nubeer
  • nubeer
you can't do this one with undetermined cofficients. you would have to use variation of parameter.
nubeer
  • nubeer
to find the yp

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anonymous
  • anonymous
Why doesn't undetermined coefficients work in this case??
nubeer
  • nubeer
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?
nubeer
  • nubeer
for "x" general form of its solution of yp would be Ax+B
anonymous
  • anonymous
Is xe^(2x) a general form?
nubeer
  • nubeer
general for have unknown cofficients.. this one is not general form.. but this has a general for , it would be (Ax+B)e^(cx)
anonymous
  • anonymous
Then how do you define ''general''??
nubeer
  • nubeer
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
anonymous
  • anonymous
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''
nubeer
  • nubeer
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
sirm3d
  • sirm3d
\(x e^{2x}\) still qualifies as general
anonymous
  • anonymous
and e^(2x) / x on the other hand is not general?
nubeer
  • nubeer
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.
anonymous
  • anonymous
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
nubeer
  • nubeer
lol well i am glad it worked out for u :)
anonymous
  • anonymous
lol Thanks!! :)

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