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

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RolyPolyBest ResponseYou've already chosen the best response.0
\[ y_c''4y_c'+4y_c=0\]\[ \lambda^24\lambda+4=0\]\[ \lambda=2, 2\]\[y_c = c_1e^{2x}+c_2xe^{2x}\] How can I guess the particular solution?
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
you can't do this one with undetermined cofficients. you would have to use variation of parameter.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Why doesn't undetermined coefficients work in this case??
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
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?
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
for "x" general form of its solution of yp would be Ax+B
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Is xe^(2x) a general form?
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
general for have unknown cofficients.. this one is not general form.. but this has a general for , it would be (Ax+B)e^(cx)
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
Then how do you define ''general''??
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
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
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
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''
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
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
 one year ago

sirm3dBest ResponseYou've already chosen the best response.0
\(x e^{2x}\) still qualifies as general
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
and e^(2x) / x on the other hand is not general?
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
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.
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.0
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
 one year ago

nubeerBest ResponseYou've already chosen the best response.1
lol well i am glad it worked out for u :)
 one year ago
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