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

Mathematics
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y_p doesn't contain constants of integration.
@experimentX , what do you mean by that?
what's the difference between solution of \( y''+y'-2y=0 \) and \( y''+y'-2y=\sin x \)

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Other answers:

one has a particular solution and the other doesn't
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
could you illustrate that? sorry :S
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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
makes sense. Thanks!
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)
honestly i would prefer "Reduction of Order" or "Annihilation Operator" method .. they are quite faster than "Undetermined coefficients"
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
well ... the second case http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients#Typical_forms_of_the_particular_integral
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 :)
sorry .. went offline :(

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