Here's the question you clicked on:
guess
find the general solution y``-5y`-6y=cos x
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.
@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:)
@RolyPoly let him solve:)
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..
y(h)=c1e^(-x)+c2e^(6x) g(x)=cos x y(P)=a cosx+b sin x >> I stood at the 5A-7B=0 -(5B+7A)=0 Then ??
@RolyPoly @Aperogalics @scarydoor thank you but i need some help
Hmm... Do you mind showing your work for the particular solution?
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/Ordinary-Differential-Equations-Dover-Mathematics/dp/0486649407/ref=sr_1_1?ie=UTF8&qid=1354192173&sr=8-1&keywords=ordinary+differential+equations It's about the best book you can get for it.
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
Of course @RolyPoly y(P)=a cosx+b sin x y`=-A sin x+b cos x y``=a cos x-b sin x y``=-y -y-5y`(-a sin x+b cos x)-6y =cos x 5asinx-5b cos x-7acosx -7b sin x=cos x (5a-7b)sinx-(5b+7a)cos x=cosx 5A-7B=0 -(5B+7A)=0
y`=-A sin x+b cos x Isn't y'' = -Acosx - Bsinx?
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 x-b sin x = -y So, -y-5y`-6y =cos x -7 y - 5y' =cos x (5a-7b)sinx-(5b+7a)cos x=cosx 5a-7b=0 -(5b+7a)=0 (it should be -(5b+7a) =1?!)
When you get the two equations: 5a-7b =0 and -(5b+7a) =1 , you can solve a and b and you'll get the particular solution.
@RolyPoly no maybe -5b-7a
(5a-7b)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 5a-7b =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?
@RolyPoly yes thank you so we get a=7/74 b=5/74 ?
You've missed a negative sign for both a and b...