Sin+cos solution undetermined coeff. de

  • Thread starter EvLer
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In summary, the conversation discusses finding a particular solution for a differential equation, specifically y'' + 16y = 24 cos(4x). The speaker has found a homogeneous solution and is working on finding the particular solution using Euler's formula. They have tried a particular solution of y = x(Acos(4x) + Bsin(4x)) and have differentiated it and plugged it into the equation. After simplifying, they are left with -8Asin(4x) + 8Bcos(4x) = 24 cos(4x) and solve for A and B. They also explore another potential particular solution of yp = x2(Acos(4x) + Bsin(4x))
  • #1
EvLer
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This looks like a very simple one, but i ran out of ideas. I do not get the answer and unlikely will get it without some help:

y'' + 16y = 24 cos(4x)

I found homogeneous solution:

y = c1ei4 + c2e-i4

I haven't re-written it using euler's formula yet.
So, for particular solution the general form is:

y = x(Acos(4x) + Bsin(4x))

I differentiated it and plugged into the DE and here's what i have:

-8Asin(4x) + 8Bcos(4x) + xsin(4x)*[16A - 16B] + xcos(4x)*[16B - 16A] = 24cos(4x);

It seems like last two terms need to go away, but then A = B and to get rid of the sin A = 0! the answer is totally different though.
Where am i going wrong?

Thanks for any hints.

EDIT: or would yp = x2(Acos(4x) + Bsin(4x))
I'm confused... :confused: i can't work it out with x2 either ...
 
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  • #2
I find the last two terms do cancel

y = x(Acos(4x) + Bsin(4x))

y' = Acos(4x) + Bsin(4x) + x (-4Asin(4x) + 4Bcos(4x))

y" = -4Asin(4x) + 4Bcos(4x) -4Asin(4x) + 4Bcos(4x) + x (-16Acos(4x) - 16Bsin(4x))

Putting back into the equation y" + 16y = 24cos(4x) gives

-4Asin(4x) + 4Bcos(4x) -4Asin(4x) + 4Bcos(4x) + x (-16Acos(4x) - 16Bsin(4x))

+ 16 ( x(Acos(4x) + Bsin(4x)) ) = 24 cos(4x)

Here we are left with

-8Asin(4x) + 8Bcos(4x) = 24 cos(4x)

so B = 3 and A = 0

so a

y(p) = 3x sin(4x)
 
  • #3
Thanks for reply.
that was my "other" solution. Good thing it matched yours. Come to find out: the book answer was wrong :mad:
 

What is the "sin+cos solution undetermined coeff. de" problem?

The "sin+cos solution undetermined coeff. de" problem is a mathematical concept that involves solving a system of equations with both sine and cosine functions as well as undetermined coefficients. This type of problem often arises in differential equations and can be solved using various techniques, such as the method of undetermined coefficients.

Why is this problem important in the field of science?

This problem is important in the field of science because it allows scientists to model and understand complex systems that involve both sine and cosine functions. These types of systems can be found in many areas of science, such as physics, engineering, and biology.

What is the method of undetermined coefficients?

The method of undetermined coefficients is a technique used to solve a system of equations with unknown coefficients. It involves assuming a solution form and then plugging it into the equations to determine the coefficients that satisfy the system. This method is often used to solve the "sin+cos solution undetermined coeff. de" problem.

What are some applications of the "sin+cos solution undetermined coeff. de" problem?

The "sin+cos solution undetermined coeff. de" problem has many applications in science and engineering. It can be used to model and analyze systems that involve oscillations, such as a pendulum or a spring-mass system. It is also useful in solving differential equations that arise in various fields, such as circuit analysis and heat transfer.

Are there any limitations to the "sin+cos solution undetermined coeff. de" problem?

Yes, there are limitations to the "sin+cos solution undetermined coeff. de" problem. This method is only applicable to linear systems, and it may not work for more complex systems that involve nonlinearities or higher order derivatives. Additionally, it may not always provide an exact solution and may require further approximations or numerical methods.

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