Question

find the general solution
y``-5y`-6y=cos x

Answer 1

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.

Answer 2

@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:)

Answer 3

@RolyPoly let him solve:)

Answer 4

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..

Answer 5

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 ??

Answer 6

@RolyPoly @Aperogalics @scarydoor thank you but i need some help

Answer 7

Hmm... Do you mind showing your work for the particular solution?

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

@scarydoor thanks

Answer 12

y`=-A sin x+b cos x
Isn't y'' = -Acosx - Bsinx?

Answer 13

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?!)

Answer 14

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.

Answer 15

@RolyPoly no maybe -5b-7a

Answer 16

(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?

Answer 17

@RolyPoly yes thank you
so we get a=7/74 b=5/74 ?

Answer 18

Not quite.

Answer 19

You've missed a negative sign for both a and b...

Answer 20

@RolyPoly thank you

Answer 21

Welcome :)