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 2 years ago
For \[y''+y'2y=sinx\]
\[y_p=Acosx+Bsinx\]
why is it incorrect to guess:
\[y_p=Asinx\]
 2 years ago
For \[y''+y'2y=sinx\] \[y_p=Acosx+Bsinx\] why is it incorrect to guess: \[y_p=Asinx\]

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experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2y_p doesn't contain constants of integration.

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0@experimentX , what do you mean by that?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2what's the difference between solution of \( y''+y'2y=0 \) and \( y''+y'2y=\sin x \)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0one has a particular solution and the other doesn't

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2for second order equation, you should have 2 constants of integration ... first one will have two constants .. now if you put another constant to particular solution ... and you will have 3 constants

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0could you illustrate that? sorry :S

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1352301416695:dw

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2guessing particular solution is quite difficult job ... easy when you have polynomials one the RHS. probably you are seeking this method http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0makes sense. Thanks!

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2the particular solution will be of the form \[ y_p=A\cos x+B\sin x \] you have to find the values of A and B by plugging into equation. I probably understand your Q quite clearly now.  plug \( y_p = A \sin (x) \) into DE, you have \[ y'' = \sin(x) \text{ and } y'(x) = \cos(x) \] no matter what the value of constant's ... there is no Cos(x) on RHS ... you cannot simply have Yp = A sin(x)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2honestly i would prefer "Reduction of Order" or "Annihilation Operator" method .. they are quite faster than "Undetermined coefficients"

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0Example 1's y_c is that in the above attachment, but how is that relevant when example 1 is y''+y'2y=x^2 ??? I guess I'll just stick with "seeing patterns" and memorization :P

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2well ... the second case http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients#Typical_forms_of_the_particular_integral

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0Oh where would I be without Paul's online notes? http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx In example 3 he does what I would have guessed y_p=Asin(2t) but then he goes on to explain why that guess is flawed. something that @experimentX and @TuringTest have been trying to explain to me all along :S Thank's guys :)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.2sorry .. went offline :(
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