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

[vish_maths]

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)

 

[mitochan]

\frac{1}{D+ia}e^{iax}=\frac{D-ia}{D^2+a^2}e^{iax}=0

 

[vish_maths]

Thanks for the reply. But, Don't we have the relation $$\dfrac {1}{f(D)}e^{cx} = \dfrac{1}{f(c)} e^{cx},~f(c) \ne 0$$

 

[mitochan]

(D+c)e^{cx}=2ce^{cx}
I do not think your equation holds.

 

[vish_maths]

(D+c)e^{cx}=2ce^{cx}
I do not think your equation holds.

Sorry I meant Don't we have the relation $$\dfrac {1}{f(D)}e^{cx} = \dfrac{1}{f(c)} e^{cx},~f(c) \ne 0$$

 

[mitochan]

De^{cx}=ce^{cx}
D^ne^{cx}=c^ne^{cx}
So
f(D)e^{cx}=f(c)e^{cx}
where polynomial f(x)
f(x)=\sum a_n x^n

So I find my previous comments wrong.

(D^2+a^2)y=0 is a equation of harmnic oscillation so it has a general solution
y=A \sin ax + B \cos ax
which can be added to the solution of your method II.

 

[PeroK]

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.