Struggling to find solution to 1D wave equation in the following form:

[Ibidy]

Homework Statement

Seperate 1D wave equation into time dependent and indipendent form and show solution takes the following trig form.

Relevant Equations

1D wave equaiton

 

[haruspex]

You should know a relationship between $e^{ix}$ and trig functions of x.

 

[Ibidy]

You should know a relationship between $e^{ix}$ and trig functions of x.

I tried using eulers identity but i just end up with a mess of complex and none complex trig functions rather than what they want.

 

[pasmith]

You want a real value for $T(t) = C_1e^{i\lambda ct} + C_2e^{-i\lambda ct}$. So $C_1 = A + iB$ and $C_2$ must be complex conjugates. Try expanding $$(A + iB)(\cos \lambda ct + i\sin \lambda ct) + (A - iB)(\cos \lambda ct - i\sin \lambda ct)$$ and see what you get.

 

[Ibidy]

Im not quite sure how or what you did but its closer to anything i was able to find.

Can i ask what you did?

 

[haruspex]

In your solution in post #1, A, B, C and D can be complex, but you know that T and $\phi$ are real. That should allow you to establish some relationships between your constants.

 

[kuruman]

I tried using eulers identity but i just end up with a mess of complex and none complex trig functions rather than what they want.

Perhaps your difficulty lies in writing the same symbols for constants that are different. Equation (15) is
$T(t)=A\sin(c\lambda t)+B\cos(c\lambda t).$ You chose to write the general solution you derived as $T(t)=Ae^{i\lambda ct}+Be^{-i\lambda ct}.$ Why did you assign $A$ to the positive exponential and $B$ to the negative instead of the other way around? You could also have chosen linear combinations of $A$ and $B$ in front of each exponential and you would still have a solution. Constants $A$ and $B$ in equation (15) do not have the same meaning as your $A$ and $B$. The same is also true for $C$ and $D$.

If I were you, I would write my general solution as $T(t)=A'e^{i\lambda ct}+B'e^{-i\lambda ct}$ and then find a relation between the primed and unprimed constants. Start with equation (15) and observe that $$\cos \!x=\frac{e^{ix}+e^{-ix}}{2}~;~~\sin \!x=\frac{e^{ix}-e^{-ix}}{2i}=-i\frac{e^{ix}-e^{-ix}}{2}.$$Then gather like exponential terms. It's not that much of a mess.

 

[robphy]

You can avoid complex numbers altogether
if you recognize that sine and cosine are solutions
of the separated 2nd-order ordinary differential equations,
since they are [also] functions that are proportional to minus-their-second-derivative
(as verified by substitution).

Otherwise, as @kuruman has noted,
you will have to work with generally distinct complex coefficients
between the different forms of the general solution.

 

[haruspex]

Perhaps your difficulty lies in writing the same symbols for constants that are different.

I think it more likely that @Ibidy understands that the two sets of A, B, C, D are different, but has missed that in general they are complex. Without that, it is not possible to transmute the one set into the other.