# Question about the constant that occures after differentiation/integration

• Dec 9th 2009, 11:26 AM
luckytommy
Question about the constant that occures after differentiation/integration
Hi!

I have a question about something I find a bit peculiar. The textbook of my math-course is a bit inconistent in the use of the constant (here $C$) that comes from differentiating an expression.

I have an expression that has been differentiated, and it looks like this:

$e^{\frac{1}{2}x^2+C}$

Sometimes this kind of expression is used to calculate further, but other times it is manipulated into this:

$Ce^{\frac{1}{2}x^2}$

My question is this:
Can I always perfrom this manipulation, or is it just in certain circumstances? I know the constant $C$ is a constant that can be changed if it is only affected by other constants (like $e$ in this situation), but the whole thing is very confusing...

Any clarifying help will be greatly appreciated!

Thomas
• Dec 9th 2009, 12:01 PM
e^(i*pi)
Quote:

Originally Posted by luckytommy
Hi!

I have a question about something I find a bit peculiar. The textbook of my math-course is a bit inconistent in the use of the constant (here $C$) that comes from differentiating an expression.

I have an expression that has been differentiated, and it looks like this:

$e^{\frac{1}{2}x^2+C}$

Sometimes this kind of expression is used to calculate further, but other times it is manipulated into this:

$Ce^{\frac{1}{2}x^2}$

My question is this:
Can I always perfrom this manipulation, or is it just in certain circumstances? I know the constant $C$ is a constant that can be changed if it is only affected by other constants (like $e$ in this situation), but the whole thing is very confusing...

Any clarifying help will be greatly appreciated!

Thomas

As e is a constant and C is a constant it follows that e^C must be a constant (much like e^2 is)

For exponential functions the laws of exponents can be used. In this case $e^{f(x)+k} = Ce^{f(x)}$ where $C = e^k$

I always denote the initial constant of integration with a different letter (usually k) and define what C is equal to at the end to avoid ambiguity
• Dec 9th 2009, 12:19 PM
luckytommy
Thank you!

In my course we have been taught to "re-use" the C, but to explain how the C has changed on the right side of the expression where the "new" C first occures.

I think we learn it this way because it seems easier (it is a beginners course).

Thanks again, I am now more confident in my management of this constant.(Clapping)
• Dec 9th 2009, 12:37 PM
chisigma
Threre is some controversials about the question. lets suppose to have the following DE...

$y^{'} -y=0$ (1)

... which can be resolved in 'standard fashion' separating the variables as follows...

$\frac{dy}{y} = dx$ (2)

Integrating both terms of (2) we have...

$\ln |y| = x + c$ (3)

... where $c$ is an 'arbitrary constant' that we can write as $c=\ln \chi$, $\chi>0$. With exponentiation of both terms of (3) we finally obtain...

$|y|= \chi \cdot e^{x}$ (4)

... that is the 'general solution' of (1). Of course the (4) can be written as...

$y= \pm \chi \cdot e^{x}$ (5)

... and it covers all the possible IVP expressed by the (1)... with the only exception of the 'initial codition' $y(0)=0$ that has as solution $y(x)=0$ that doesn't correspond to any value of $\chi$ because we have supposed $\chi >0$ (Thinking)...

http://digilander.libero.it/luposabatini/notte1.jpg

Merry Christmas from Italy

$\chi$ $\sigma$