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<math>(e^z f)''=e^zf +2e^zf'+e^zf''=2(e^zf + e^zf')=2(e^zf)'.</math>
 
<math>(e^z f)''=e^zf +2e^zf'+e^zf''=2(e^zf + e^zf')=2(e^zf)'.</math>
  
Let <math>g=e^zf.</math>  Then g'=2g.
+
Let <math>g=e^zf.</math>  Then <math>g'=2g</math> and now you can
 +
use the theorem from class that concerns solutions of this first
 +
order complex ODE.  By the way, you will also need to use the fact
 +
that if two analytic functions on the complex plane have the same
 +
derivative, then they must differ by a constant.

Revision as of 07:42, 23 September 2009


Homework 4

HWK 4 problems

Hint for IV.6.3 --Steve Bell

We assume $ f''=f $ on $ \mathbb C $.

Notice that

$ (e^z f)''=e^zf +2e^zf'+e^zf''=2(e^zf + e^zf')=2(e^zf)'. $

Let $ g=e^zf. $ Then $ g'=2g $ and now you can use the theorem from class that concerns solutions of this first order complex ODE. By the way, you will also need to use the fact that if two analytic functions on the complex plane have the same derivative, then they must differ by a constant.

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