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scarydoorBest ResponseYou've already chosen the best response.0
this is a 2nd order linear differential equation. There is a really easy method for solving these things. Do you know of it? You should have been told it in class. Otherwise you can search google and it'll tell you the method.
 one year ago

AperogalicsBest ResponseYou've already chosen the best response.0
@guess it is a second order equation it as two parts one general and particular and their sum is the complete solution so for general put y''5y'6y=0 and replace y^n by D^n and solve quadratic...............try it:)
 one year ago

AperogalicsBest ResponseYou've already chosen the best response.0
@RolyPoly let him solve:)
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.2
First, find the complementary solution by setting the left side equal to 0. y``5y`6y=0 < homogeneous solution, so, λ^2  5λ 6 =0 Solve λ \[y_c = c_1e^{\lambda _1 x} + c_2e^{\lambda _2 x}\] Then, find out the particular solution, try \(y_p = Asinx+Bcosx\) (No guarantee that it works though). Differentiate it and solve the A and B. Then, you get \(y_p\) \[y = y_c +y_p\] @Aperogalics I never intend to give out the answer..
 one year ago

guessBest ResponseYou've already chosen the best response.0
y(h)=c1e^(x)+c2e^(6x) g(x)=cos x y(P)=a cosx+b sin x >> I stood at the 5A7B=0 (5B+7A)=0 Then ??
 one year ago

guessBest ResponseYou've already chosen the best response.0
@RolyPoly @Aperogalics @scarydoor thank you but i need some help
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.2
Hmm... Do you mind showing your work for the particular solution?
 one year ago

scarydoorBest ResponseYou've already chosen the best response.0
It's called 'method of undetermined coefficients'. If you're going to be doing differential equations a fair bit, then I recommend this book: http://www.amazon.com/OrdinaryDifferentialEquationsDoverMathematics/dp/0486649407/ref=sr_1_1?ie=UTF8&qid=1354192173&sr=81&keywords=ordinary+differential+equations It's about the best book you can get for it.
 one year ago

scarydoorBest ResponseYou've already chosen the best response.0
Do you know how it works? The method to find a particular solution? That seems to be what you're stuck on. This is the method to use: http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients
 one year ago

guessBest ResponseYou've already chosen the best response.0
Of course @RolyPoly y(P)=a cosx+b sin x y`=A sin x+b cos x y``=a cos xb sin x y``=y y5y`(a sin x+b cos x)6y =cos x 5asinx5b cos x7acosx 7b sin x=cos x (5a7b)sinx(5b+7a)cos x=cosx 5A7B=0 (5B+7A)=0
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.2
y`=A sin x+b cos x Isn't y'' = Acosx  Bsinx?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.2
But that doesn't affect your conclusion that y'' = y y(P)=a cosx+b sin x y`=A sin x+b cos x y``=a cos xb sin x = y So, y5y`6y =cos x 7 y  5y' =cos x (5a7b)sinx(5b+7a)cos x=cosx 5a7b=0 (5b+7a)=0 (it should be (5b+7a) =1?!)
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.2
When you get the two equations: 5a7b =0 and (5b+7a) =1 , you can solve a and b and you'll get the particular solution.
 one year ago

guessBest ResponseYou've already chosen the best response.0
@RolyPoly no maybe 5b7a
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.2
(5a7b)sinx(5b+7a)cos x=cosx Compare the terms on the left and right, there is no sinx term, so the coefficient of sinx is 0 Hence you get the first equation 5a7b =0 there is a cosx term, with the coefficient 1, so the coefficient of cosx term on the left is equal to 1. Hence we should get (5b+7a) = 1 as the second equation. Got it?
 one year ago

guessBest ResponseYou've already chosen the best response.0
@RolyPoly yes thank you so we get a=7/74 b=5/74 ?
 one year ago

RolyPolyBest ResponseYou've already chosen the best response.2
You've missed a negative sign for both a and b...
 one year ago
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