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.