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

In summary: I think that should help you to see why there can't be another solution.In summary, based on the given data and the attempt at solving the problem, the function f(x) is a constant function with f(x) = 0 being the only solution that satisfies the given differential equation and initial condition. The uniqueness of the solution further supports this conclusion.
  • #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
 
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  • #2
ritwik06 said:

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
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.
 
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  • #4
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
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
 
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  • #6
ritwik06 said:

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

What is a constant function?

A constant function is a type of function in mathematics where the output value is the same for every input value. This means that no matter what value is plugged into the function, the result will always be the same.

What does it mean for f'(x) to equal f(x)?

When f'(x) is equal to f(x), it means that the derivative of the function is equal to the function itself. In other words, the slope of the function at any point is equal to the value of the function at that point.

How do I solve a constant function homework with f'(x)=f(x) and f(0)=0?

To solve this type of homework, you can use the fact that f'(x) is equal to f(x) to rewrite the function as f'(x)=f(x). Then, you can use the initial condition f(0)=0 to solve for the constant value of the function. This will give you the complete solution to the problem.

Why is f(0)=0 important in solving this type of function?

The initial condition f(0)=0 is important because it gives us a specific point on the function to work with. This allows us to solve for the constant value of the function and find the complete solution.

What are some real-life applications of constant functions?

Constant functions have many real-life applications, such as in physics to represent the velocity of a stationary object, in economics to model a fixed cost, and in chemistry to describe an unchanging reaction rate. They are also commonly used in computer programming to define constants that do not change throughout the program.

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