# A silly question regarding general solutions to second order DE's

It's been a while since I've had DE, and it seems I have forgotten something basic, yet crucial. In solving the Time-Independent Schrodinger Wave Equation in one dimension, I have the following:

$$\left( -\frac{\hbar}{2m} \frac{\partial^2}{\partial x^2}+V\right) \Psi(x)=E\Psi(X)$$

$$\frac{\partial^2}{\partial x^2}\Psi(x)+K^2\Psi(x)=0$$

where V=0 and

$$K^2=\frac{2mE}{\hbar^2}$$

And my question is: are the following solutions equal?

$$\Psi(x)=C_1e^{iKx}+C_2e^{-iKx}$$

$$\Psi(x)=C_1Cos(Kx)+C_2Sin(Kx)$$

I know that Euler's Formula is $$e^{i\theta x}=Cos(\theta x)+iSin(\theta x)$$, however inserting this into the first solution above does not result in the second. Thanks!

IHateMayonnaise

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Dick
Homework Helper
There is no difference between those two forms of the solution. But they aren't 'equal'. When you go from one to the other you need to change the values of C1 and C2. For example C1 in the real form is equal to C1+C2 in the complex form, C2 in the real form is i*(C1-C2) in the complex form.

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There is no difference between those two forms of the solution. But they aren't 'equal'. When go from one to the other you need to change the values of C1 and C2. For example C1 in the real form is equal to C1+C2 in the complex form.
Well, when solving The Schrodinger equation, how do I know which form to take? Can I just put it in one form or the other and solve for the coefficients?

Dick
Homework Helper
You can use them interchangeably. E.g. 2*exp(iKx)+3*exp(-iKx)=(2+3)*cos(Kx)+i*(2-3)*Sin(Kx) (check that using Euler's formula). They are really just plain equal. Depends on whether you have a preference for where to put the i's.

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You can use them interchangeably. E.g. 2*exp(iKx)+3*exp(-iKx)=(2+5)*cos(Kx)+i*(2-3)*Sin(Kx) (check that using Euler's formula). They are really just plain equal. Depends on whether you have a preference for where to put the i's.
Well what you have there makes sense to me, what confuses me is why there isn't an i in front of the sine term in the second solution

$$\Psi(x)=C_1e^{iKx}+C_2e^{-iKx}$$

$$\Psi(x)=C_1Cos(Kx)+C_2Sin(Kx)$$

Since

$$\Psi(x)=C_1e^{iKx}+C_2e^{-iKx}=C_1(Cos(Kx)+iSin(Kx))+C_2(Cos(Kx)-iSin(Kx)) = (C_1+C_2)Cos(Kx)+(C_1-C_2)iSin(Kx)$$

and the "i" is still there!! Why?? (I get the part about the coefficients)

Dick