Constant Function

1. Jun 1, 2008

ritwik06

1. The problem statement, all variables and given/known data
Data Given:
f'(x)=f(x) ..........(i)
and
f(0)=0 ..........(ii)

What kind of function is f(x)?

3. The attempt at a solution
From (i)
dy/dx=y
1/y dy= dx
Integrating;
ln y = x+C
From (ii)
ln 0=0+c
Therefore; c is not defined!

My book gives the answer that f(x) is a constant function. But how is it true? Is my way of solving wrong?
regards,
Ritwik

2. Jun 1, 2008

rock.freak667

From the part in bold

y=exp(x+c)

3. Jun 1, 2008

Vid

You divided by y when y = 0. I don't think you're supposed to solve it, but rather argue why it can't be anything other than a constant function based on the equation and the initial values.

Last edited: Jun 1, 2008
4. Jun 1, 2008

Raze2dust

y=any constant clearly doesn't satisfy (i)

though y=0 does satisfy..i think the answer is y=0 not y=any constant

when you do dy/y=dx, you are assuming that y is not equal to zero, because dy/y will not be defined for y=0

So, you have to consider the case y=0 separately

5. Jun 1, 2008

GregA

The solution of your ODE should give y = Ae^x (for some constant A)
Applying your initial condition gives 0 = Ae^0 ==> A = 0

If I'm right the function y = 0 is just as much a constant function as say the function y = 5

Last edited: Jun 1, 2008
6. Jun 2, 2008

HallsofIvy

Staff Emeritus
No, integrating does NOT give "ln y= x+ C". Integrating gives ln |y|= x+ C. That's your crucial error.

To see why that is important, take the exponential of both sides.
eln |y|= ex+ C or |y|= C' ex where C'= eC. While eC must be positive, we can drop the absolute value by allowing C' to be negative as well, and, by continuity, C'= 0.

y(x)= C'ex obviously satisifies the given differential equation for any real number C'.

Now, put y=0, x= 0 into y= C'ex. What is C'? What is y(x)?

7. Jun 2, 2008

spideyunlimit

Data Given:
f'(x)=f(x) ..........(i)
and
f(0)=0 ..........(ii)

What kind of function is f(x)?
-------------------------

f(x) is a constant function..... f(x) = 0
where f'(x) = f(x) holds true. (The only other possibility was f(x) = e^x but then f(0) is not 0.)

8. Jun 2, 2008

spideyunlimit

From your attempt at the solution you just get that c = 0.... but that doesn't help really.

9. Jun 2, 2008

nicksauce

There is a simple solution f(x) = 0. What do you know about the uniqueness of the solution?