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Calculating the value of 1/{D^2+a^2} sin ax

  • Thread starter vish_maths
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  • #1
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Homework Statement:

I have tried calculating using two different methods and both of them yield different results. Could someone please pinpoint any error in Method 1 which I might be making. **My textbook uses method 2.** Thanks a lot for your help

Relevant Equations:

1/(D-b) e^(bx) = x e^(bx)
WhatsApp Image 2020-02-03 at 4.58.15 PM.jpeg
WhatsApp Image 2020-02-03 at 4.58.15 PM (1).jpeg
 

Answers and Replies

  • #2
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[tex]\frac{1}{D+ia}e^{iax}=\frac{D-ia}{D^2+a^2}e^{iax}=0[/tex]
 
  • #3
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Thanks for the reply. But, Don't we have the relation $$\dfrac {1}{f(D)}e^{cx} = \dfrac{1}{f(c)} e^{cx},~f(c) \ne 0$$
 
  • #4
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[tex](D+c)e^{cx}=2ce^{cx}[/tex]
I do not think your equation holds.
 
  • #5
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[tex](D+c)e^{cx}=2ce^{cx}[/tex]
I do not think your equation holds.
Sorry I meant Don't we have the relation $$\dfrac {1}{f(D)}e^{cx} = \dfrac{1}{f(c)} e^{cx},~f(c) \ne 0$$
 
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  • #6
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[tex]De^{cx}=ce^{cx}[/tex]
[tex]D^ne^{cx}=c^ne^{cx}[/tex]
So
[tex]f(D)e^{cx}=f(c)e^{cx}[/tex]
where polynomial f(x)
[tex]f(x)=\sum a_n x^n[/tex]

So I find my previous comments wrong.

[tex](D^2+a^2)y=0[/tex] is a equation of harmnic oscillation so it has a general solution
[tex]y=A \sin ax + B \cos ax[/tex]
which can be added to the solution of your method II.
 
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  • #7
PeroK
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Homework Statement:: I have tried calculating using two different methods and both of them yield different results. Could someone please pinpoint any error in Method 1 which I might be making. **My textbook uses method 2.** Thanks a lot for your help
Relevant Equations:: 1/(D-b) e^(bx) = x e^(bx)

View attachment 256543View attachment 256544
Why do you think you've done anything wrong? Have you checked your answer?

I must admit I'm not familiar with this technique, but you have picked up an additional solution to the homogeneous equation.

Given this is a second-order DE, there ought to be two linearly independent solutions.

You have them both; the book method finds only one.
 
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