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

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anonymous
 4 years ago
Best 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?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why doesn't undetermined coefficients work in this case??

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.1well 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
 4 years ago
Best ResponseYou've already chosen the best response.1for "x" general form of its solution of yp would be Ax+B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is xe^(2x) a general form?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then how do you define ''general''??

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.1like 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
 4 years ago
Best ResponseYou've already chosen the best response.0Some 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
 4 years ago
Best ResponseYou've already chosen the best response.1in 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
 4 years ago
Best ResponseYou've already chosen the best response.0\(x e^{2x}\) still qualifies as general

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and e^(2x) / x on the other hand is not general?

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.1in 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
 4 years ago
Best ResponseYou've already chosen the best response.0Variation 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
 4 years ago
Best ResponseYou've already chosen the best response.1lol well i am glad it worked out for u :)
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