First order linear D.E. involving i

Syrus
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Homework Statement




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?


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The Attempt at a Solution

 
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Syrus said:

Homework Statement

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?

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}##?
 
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.
 
Syrus said:
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.

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

What's the integrating factor here?
 
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