# First order linear D.E. involving i

1. Mar 11, 2013

### Syrus

1. The problem statement, all variables and given/known data

I am familiar with the standard method of obtaining a solution to a first-order, linear D.E. (i.e. using an integrating factor). However, consider the D.E. if'(x) = qf(x). It seems (after recasting the equation in the proper form) the solution suggested by the above method is f(x) = exp(iqx). However, this solution does not satisfy the original D.E. The correct solution seems to be exp(-iqx). I am curious as to why the method isn't working here. Furthermore, what general branch(s) deals with such topics if not general O.D.E.s?

2. Relevant equations

3. The attempt at a solution

2. Mar 11, 2013

### Curious3141

If q is a (complex) constant, and i represents the square root of -1, then this is just a simple separable ODE.

$if'(x) = qf(x)$

$\frac{f'(x)}{f(x)} = \frac{q}{i} = -iq$

Integrating both sides wrt x,

$\ln(f(x)) = -iqx + \ln C$

$f(x) = Ce^{-iqx}$

where C is an arbitrary (complex) constant.

So what did you do to get the answer involving $e^{iqx}$?

3. Mar 11, 2013

### Syrus

Ah, I see. I was using an integrating factor:

That is, for equations of form

y'(x) + h(x)y(x) = g(x)

obtaining a solution y = y(x) via y = exp(∫h(x)dx).

But this doesn't seem to yield the same answer.

4. Mar 12, 2013

### Curious3141

Sure it does. Go through the working more carefully (you should show it here).

What's the integrating factor here?

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