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Constant Function

  • Thread starter ritwik06
  • Start date
  • #1
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Homework Statement


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

What kind of function is f(x)?



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?
Please Help
regards,
Ritwik
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31

Homework Statement


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

What kind of function is f(x)?



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!
From the part in bold

y=exp(x+c)
then use your initial values.
 
  • #3
Vid
401
0
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:
  • #4
63
0
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
210
0
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:
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,793
922

Homework Statement


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

What kind of function is f(x)?



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!
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'.

My book gives the answer that f(x) is a constant function. But how is it true? Is my way of solving wrong?
Please Help
regards,
Ritwik
Now, put y=0, x= 0 into y= C'ex. What is C'? What is y(x)?
 
  • #7
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
From your attempt at the solution you just get that c = 0.... but that doesn't help really.
 
  • #9
nicksauce
Science Advisor
Homework Helper
1,272
5
There is a simple solution f(x) = 0. What do you know about the uniqueness of the solution?
 

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