Undetermined coefficient (y'' + 4y) = (x^2) +2

[werson tan]

Homework Statement

i gt stucked here ... how to continue ? what's wrong with my working ? i found 2 values for D , whcih is wrong

Homework Equations

The Attempt at a Solution

 

[SteamKing]

Homework Statement

i gt stucked here ... how to continue ? what's wrong with my working ? i found 2 values for D , whcih is wrong

Homework Equations

The Attempt at a Solution

Your selection of yp is confusing. It is recommended that if the RHS is a polynomial of order n, then yp = a general polynomial of order n, not order n+1.

See:

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

 

[werson tan]

Your selection of yp is confusing. It is recommended that if the RHS is a polynomial of order n, then yp = a general polynomial of order n, not order n+1.

See:

http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx

then , what is the correct yp ?

i choose my initial yp = cX^2 + Dx + E , nut i found that the term E is similar to the A , so i multiply the initial yp with x , to get my new yp

 

[SteamKing]

then , what is the correct yp ?

i choose my initial yp = cX^2 + Dx + E , nut i found that the term E is similar to the A , so i multiply the initial yp with x , to get my new yp

A and E are solitary constants. You don't need to multiply your initial choice of yp by x.

 

[HallsofIvy]

Your difficulty is that your solution to the associated homogeneous equation is wrong! The associated homogeneous equation is y''+ 4y= 0 which has characteristic equation $m^2+ 4= 0$ with solutions 2i and -2i. You have written the characteristic equation as $m^2+ 4m= 0$ which has solutions m= -4 and m= 0. That would correspond to differential equation y''+ 4y'= 0.

 

[werson tan]

Your difficulty is that your solution to the associated homogeneous equation is wrong! The associated homogeneous equation is y''+ 4y= 0 which has characteristic equation $m^2+ 4= 0$ with solutions 2i and -2i. You have written the characteristic equation as $m^2+ 4m= 0$ which has solutions m= -4 and m= 0. That would correspond to differential equation y''+ 4y'= 0.

i have redo the question , i got Acos2x + Bsin2x +0.25x^2 + 3/2 , but the ans given is Acos2x + Bsin2x +0.25x^2 +3/8

 

[SteamKing]

i have redo the question , i got Acos2x + Bsin2x +0.25x^2 + 3/2 , but the ans given is Acos2x + Bsin2x +0.25x^2 +3/8

Check your arithmetic or provide your work to obtain this solution.

 

[werson tan]

Check your arithmetic or provide your work to obtain this solution.

i have attached the working in the picture uploaded above

 

[Mark44]

i have redo the question , i got Acos2x + Bsin2x +0.25x^2 + 3/2 , but the ans given is Acos2x + Bsin2x +0.25x^2 +3/8

I deleted the picture you attached. It was extremely hard to read, with the top so dark it was nearly impossible to read, and portions were cut off.

i have attached the working in the picture uploaded above

It's gone now. We would prefer that you show your work here in the input pane, rather than in a poor quality image with bad lighting and information cut off.

I should add that you're 1 point away from a permanent ban. You should be extra careful to make sure that your posts follow the rules here.

 

[Mark44]

i gt stucked here

"Stucked" is not a word in English. You can get stuck or you got stuck, but you can't get "stucked."

 

[epenguin]

"Stucked" is not a word in English. You can get stuck or you got stuck, but you can't get "stucked."

Whatever it is you can't gt it anyway.

 

[epenguin]

And anyway people don't need to post the dark images we see a lot, there are apps like DocScan HD for iPad which clean them up. But anyway this is only recommended for diagrams - we don't like it for maths calculation pages in fact you have been warned so better not this type of work.

 

[epenguin]

If this question is meant as an in exercise in variation of parameters even though the equation can be solved in other ways then OK. But if the question is just solving the given equation then In the spirit of another post on d.e.'s I made yesterday #4 for this one too you can practically eliminate guessing and having to know stuff and you don't need variation of parameters.

The equation to be solved is

(D² + 4)y = 2 + x²

What is the RHS (D² + 4) of?

Well (D² + 4)x² = 2 + 4x²

To get something like the RHS

(D² + 4)(x²/4) = ½ + x² = (2 + x²) - 3/2

So the d.e. can be written:

(D² + 4)(y - x²/4) = - 3/2

It is not hard to find what a constant is (D² + 4) of.

-3/2 = (D² + 4)(-⅜)

So our equation is

(D² + 4)(y -x²/4 + ⅜) = 0

In other words we have transformed by a systematic procedure our original non-homogeneous equation into a homogeneous (and familiar!) homogeneous one that we know how to solve.

d²Y/dx² = -4Y

in a new variable Y

Y = ( y -x²/4 + ⅜)

I have not completed this, but we are getting the same x²/4 and ⅜ that were obtained in the other approaches, so it is looking OK.

I think the equation of #4 can also be solved this way