Solving Constant Function Homework w/ f'(x)=f(x) & f(0)=0

[ritwik06]

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

 

[rock.freak667]

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.

 

[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.

 

[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

 

[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

 

[HallsofIvy]

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.
e^{ln |y|}= e^{x+ C} or |y|= C' e^x where C'= e^C. While e^C 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'e^x. What is C'? What is y(x)?

 

[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.)

 

[spideyunlimit]

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

 

[nicksauce]

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