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Finding general solution to a second order forced diff eq

  • Thread starter ollyfinn
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  • #1
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I am trying to solve the following problem and am a bit lost so any advice would be welcomed.

x'' = 2x' + x = 3cos2t + sin2t

My understanding is that I need to find the general solution for the unforced equation and a particular solution of the above equation. When these are added together then I will get my final solution.

Using the charateristic equation:

m^2 + 2m + 1 = 0

And then the quadratic equation will give me:

m = -1

This gives a general solution of:

x(t) = Ae^m0t + Bte^m0t

so

Ae^-1t + Bte^-1t + particular solution = final answer

Is a particular solution

x(t) = p cos βt + q sin βt....???

Not really sure where to go from here.
 

Answers and Replies

  • #2
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I have been looking at this problem again and have come to a solution of:

x(t) = Ae^-1t + Bte^-1t + 3/4t sin 2t - 1/4t cos 2t

If anybody thinks this is correct then let me know. Thanks.
 
  • #3
ehild
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I am trying to solve the following problem and am a bit lost so any advice would be welcomed.

x'' = 2x' + x = 3cos2t + sin2t
Is the first '=' a minus?


ehild
 
  • #4
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Hi ehild,

My typing mistake, it should read:

x'' + 2x' + x = 3cost2t + sin2t
 
  • #5
ehild
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Your method is correct, but there must be some mistake in the derivation of the particular solution, the coefficients are wrong. Check your calculation or show it in detail.


ehild
 
  • #6
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ehild

This was the method used to get the solution I came to:

Find the general solution of

x'' + 2x' + x = 3cos2t + sin2t

Corresponding unforced

x'' + 2x' + x = 0

Using the characteristic equation

m^2 + 2m + 1 = 0

m = -1

So the corresponding equation will be

xc(t) = Ae^m0T + Bte^m0t

Xc(t) = Ae^-1t + Bte^-1t

For the particular solution

xp(t) = p cos β(t) + q sin β(t)

For 3cos2t

x(t) = p/(2β) t sin βt

x(t) = 3/4 t sin 2t

For sin 2t

x(t) = -q/(2β) t cos βt

x(t) = -1/4 t cos 2t

So particular solution is

3/4 t sin 2t - 1/4 t cos 2t

Gen sol = corr sol + part sol

Gen sol = Ae^-1t + Bte^-1t + 3/4 t sin 2t - 1/4 t cos 2t
 
  • #7
SteamKing
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In the solution to your characteristic equation, the signs of all the terms are positive, so m cannot equal -1.
 
  • #8
ehild
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In the solution to your characteristic equation, the signs of all the terms are positive, so m cannot equal -1.
??? The characteristic equation is m2+2m+1=0, that is, (m+1)2=0, m=-1.

ehild
 
  • #9
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ehild,

Can you see where I may have gone wrong in my method then?

Cheers
 
  • #10
HallsofIvy
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In the solution to your characteristic equation, the signs of all the terms are positive, so m cannot equal -1.
No, coefficents of all terms positive means there cannot be any positive roots. Certainly [itex](-1)^2+ 2(-1)+ 1= 0[/itex].

The fact that -1 is a double root means that the general solution to the associated homogeneous equation is [itex]Ce^{-t}+ Dte^{-t}[/itex].

Now, look for a particular solution of the form x= Acos(2t)+ B sin(2t). (NOT "[itex]\beta t[/itex]"- the coefficient of t in both trig functions is "2", use that.)

(If, for example, the right side had been "2cos(2t)+ sin(3t)", you would have needed to try "x(t)= Acos(2t)+ Bsin(2t)+ Ccos(3t)+ Dsin(3t)".

Then x'= -2Asin(2t)+ 2Bcos(2t), and x''= -4Acos(2t)- 4Bsin(2t)

x''+ 2x'+ x= -4Acos(2t)- 4Bsin(2t)- 4Asin(2t)+ 4Bcos(2t)+ Acos(2t)+ Bsin(2t)
= (-4A+ 4B+ A)cos(2t)+ (-4B- 4A+ B)sin(2t)= 3kcos(2t)+ sin(2t)
so -4A+4B+ A= 3, -4B- 4A+ B= 1.
 
  • #11
ehild
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For the particular solution

xp(t) = p cos β(t) + q sin β(t)

For 3cos2t

x(t) = p/(2β) t sin βt

x(t) = 3/4 t sin 2t

For sin 2t

x(t) = -q/(2β) t cos βt

x(t) = -1/4 t cos 2t

So particular solution is

3/4 t sin 2t - 1/4 t cos 2t

Gen sol = corr sol + part sol

Gen sol = Ae^-1t + Bte^-1t + 3/4 t sin 2t - 1/4 t cos 2t

I can not imagine what you did and why.

Try the particular solution in the form

xp=pcos(2t)+qsin(2t).

You need to determine x' and x'', substitute back into the original equation, collecting terms with cos(2t) and sin(2t) and equal them with the corresponding terms in the right-hand side.

And you can always check your solution by plugging back.

I just have noticed that HallsofIvy have done most of the work for you.

ehild
 
Last edited:
  • #12
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No, coefficents of all terms positive means there cannot be any positive roots. Certainly [itex](-1)^2+ 2(-1)+ 1= 0[/itex].

The fact that -1 is a double root means that the general solution to the associated homogeneous equation is [itex]Ce^{-t}+ Dte^{-t}[/itex].

Now, look for a particular solution of the form x= Acos(2t)+ B sin(2t). (NOT "[itex]\beta t[/itex]"- the coefficient of t in both trig functions is "2", use that.)

(If, for example, the right side had been "2cos(2t)+ sin(3t)", you would have needed to try "x(t)= Acos(2t)+ Bsin(2t)+ Ccos(3t)+ Dsin(3t)".

Then x'= -2Asin(2t)+ 2Bcos(2t), and x''= -4Acos(2t)- 4Bsin(2t)

x''+ 2x'+ x= -4Acos(2t)- 4Bsin(2t)- 4Asin(2t)+ 4Bcos(2t)+ Acos(2t)+ Bsin(2t)
= (-4A+ 4B+ A)cos(2t)+ (-4B- 4A+ B)sin(2t)= 3kcos(2t)+ sin(2t)
so -4A+4B+ A= 3, -4B- 4A+ B= 1.

Thanks for your help.

Am I correct in thinking that values for A and B should be solved using simultaneous equations? Then they will form part of particular solution part of the final answer?

Sorry if I am just not getting it. I think I may be well out of my depth at this level of study and I just cant get my head around the lessons I am reading through in my textbook. I really appreciate the assistance.
 
  • #13
ehild
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The particular solution must be solution of the original equation. Yours is not. Plug back x=3/4 t sin 2t - 1/4 t cos 2t and you will see.

A and B are constants, which are determined by the initial conditions.

There are two methods to find a particular solution of the inhomogeneous equation if you know the solution of the homogeneous one. One is the Method of Undetermined Coefficients, see http://www.google.com/url?sa=t&source=web&cd=5&ved=0CEIQFjAE&url=http://www.efunda.com/math/ode/linearode_undeterminedcoeff.cfm&ei=W4yITvqqLs_ysgaWsfjgAQ&usg=AFQjCNFr4f3yx1-J3B3PGXB7Ib-eTke0vQ, for example.
The other method is Variation of Parameters. http://en.wikipedia.org/wiki/Variation_of_parameters. In this case you consider A and B as functions of the independent variable, and find a particular solution in the form Xp=A(t)Xh1+B(t)Xh2.

ehild
 

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