# Math Help - points in proof of homogeneous DE

1. ## points in proof of homogeneous DE

I have a question about the proof of this elementary DE, the calculations are easy
$\frac{dy}{dt}+a(t)y=0\rightarrow \left | y(t)\right|=exp(-\int a(t)dt+ca)\rightarrow |cexp(-\int a(t)dt)|=c$
-not sure if you can add the integral into the absolute value
$y(t)exp(\int a(t)dt)$ is a continuous function of time
-I'm not sure how this is assumed to be continuous in time?
If the absolute value of a function g(t) is constant then g must be constant,
Proof:
If g isn't constant there exists t1,t2 for which g(t1)=c, g(t2)=-c. By the intermediate value theorem g must achieve all values between -c and +c which is impossible if |g(t)|=c
-but if g(t)=x or d then why does it necessarily equal c and -c

2. ## Re: points in proof of homogeneous DE

I misposted the equation it should read
$\frac{dy}{dt}+a(t)y=0\rightarrow \left | y(t)\right|=exp(-\int a(t)dt+c)\rightarrow |y(t)exp(-\int a(t)dt)|=c$

3. ## Re: points in proof of homogeneous DE

-I'm not sure how this is assumed to be continuous in time?
You are given that dy/dt exists, y is differentiable so continuous. Any integral is continuous, the exponential of a continuous function is continuous, and the product of two continuous functions is continuous. That's why $y(t)exp(\int a(t)dt)$ is continuous.

If the absolute value of a function g(t) is constant then g must be constant
No, that's not true. The absolute value of "g(t)= 1 if t is rational, g(t)= -1 if t is irrational" is the constant 1.

If g isn't constant there exists t1,t2 for which g(t1)=c, g(t2)=-c.
Again, no. The function g(x)= 1 if x is rational, 2 if x is irrational has NO pair t1 and t2 such that g(t1)= c, g(t2)= -c.

By the intermediate value theorem g must achieve all values between -c and +c which is impossible if |g(t)|=c
-but if g(t)=x or d then why does it necessarily equal c and -c
Are you assuming that g is continuous? You didn't say that.

It is true that if the absolute value of a continuous function is constant, then the function is constant.
(But "If g isn't constant there exists t1,t2 for which g(t1)=c, g(t2)=-c." is not necessarily true even for continuous functions.)

4. ## Re: points in proof of homogeneous DE

then this is an error in the Braun Springer text in the early chapters. I looked at this a LONG time but instantly thought how can he come up with g(t1)=c, g(t2)=-c from assuming g isn't constant. is this a really bad book? it covers a lot of nice applications and has some interesting problems

it bothered me so much I drank a beer and vomited. how can there be an error in this text????

5. ## Re: points in proof of homogeneous DE

I mean you have a transcendental like y=sin(x)+1 and no value of y(t)=-c

6. ## Re: points in proof of homogeneous DE

what is Braun saying here in the early chapters?