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JenniferSmart1
Group Title
For \[y''+y'2y=sinx\]
\[y_p=Acosx+Bsinx\]
why is it incorrect to guess:
\[y_p=Asinx\]
 one year ago
 one year ago
JenniferSmart1 Group Title
For \[y''+y'2y=sinx\] \[y_p=Acosx+Bsinx\] why is it incorrect to guess: \[y_p=Asinx\]
 one year ago
 one year ago

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experimentX Group TitleBest ResponseYou've already chosen the best response.2
y_p doesn't contain constants of integration.
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
@experimentX , what do you mean by that?
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
one has a particular solution and the other doesn't
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
for 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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
could you illustrate that? sorry :S
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
dw:1352301416695:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
guessing 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
 one year ago

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
makes sense. Thanks!
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
the 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)
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Example 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
 one year ago

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

JenniferSmart1 Group TitleBest ResponseYou've already chosen the best response.0
Oh 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 :)
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.2
sorry .. went offline :(
 one year ago
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