Calculus Help: Solving Homework Statement

[Matt1234]

Homework Statement

Hi guys, i need some help with the following:

[PLAIN]http://img534.imageshack.us/img534/3079/22742169.jpg

I have 2 questions:

1) why do we make it +/- e^c without specifying a domain for x?
(this is what i have done in the past for absolute value problems)


2) The part where it says we can easily see y= 0 is a solution, i don't see that at all. since e^x != 0

 

[Unit]

1) I'm not exactly sure what you mean. We do not specify a domain for x because the absolute value bars are on the variable "y". It would be the range that we should specify; the range of |y| is [0, ∞) and the range of y is R.

2) Sometimes, you'll see differential equations with multiple solutions that are different from each other. Here, y = 0 is a trivial solution (but still valid) because if y = 0 for all x, then dy/dx = 0 = 0 x^2 = y x^2. So y = 0 satisfies the original equation.

 

[Matt1234]

Thank you for your help. So with regards to number 2) the zero solution was included in the original DE that's why I must make it a soln in my final equation? It seems confusing that they state y != 0 in the line where we have dy/y.


As far as 1) goes I'm still slightly confused. You mentioned the range of y is [0, infinity) then you said the range of y Is R. This is exactly where stuck i agree that the range starts from zero but done see where the -e comes in? Does this all go back to the fact that negative numbers are solns to the original DE?



Thank you for your time,
Matt

 

[HallsofIvy]

Thank you for your help. So with regards to number 2) the zero solution was included in the original DE that's why I must make it a soln in my final equation? It seems confusing that they state y != 0 in the line where we have dy/y.


As far as 1) goes I'm still slightly confused. You mentioned the range of y is [0, infinity)

No, he didn't. You are not reading carefully. He said the range of |y| is [0, infinity).

then you said the range of y Is R. This is exactly where stuck i agree that the range starts from zero but done see where the -e comes in?

Why "-e" are you talking about? It goes back to the fact that knowing |x| does not tell us anything about the sign of x itself and so it might be either positive or negative.

Does this all go back to the fact that negative numbers are solns to the original DE?

I don't know what you mean by this. Functions are solutions to differential equations, not numbers.

Thank you for your time,
Matt

In your first post you said

2) The part where it says we can easily see y= 0 is a solution, i don't see that at all. since e^x != 0

No, but "C", the constant multiplying the exponential, can be. But the fact is that, in some cases, there may be solutions to differential equations that do NOT fit whatever "general" solution you have written. In any case, whether a function is a solution to the equation does not depend upon whether it is or is not of a particular form but whether or not it satisfies the equation. You certainly should be able to see that if y is identically equal to 0, then it is a constant, so its derivative is 0. That is, it satisfies y'= 0= x^2(0).

 

[Amok]

Dude, what is the statement:

" |y| = e^C e^{(x^3/3)}

is an implicit solution to the differential equation" telling you?

It is telling you that if

y > 0

then

y = e^C e^{(x^3/3)}

is a possible explicit solution to this equation. This can be written as

y = A e^{(x^3/3)}

with

A > 0.

The statement is also telling you that if

y < 0

then

y = -e^C e^{(x^3/3)}

is another solution. This can be written as

y = A e^{(x^3/3)},

with

A < 0.

So you can write the solution as

y = A e^{(x^3/3)}

with A non-zero real number.

To get to the previous solutions you made the assumption that

y \neq 0.

What happens if

y = 0?

You can check that this is a possible solution to the equation (trivial solution, the derivative of 0 is 0). Which means you can "summarize" these three possible cases by writing the solution as

y = A e^{(x^3/3)}
with

A \in \mathbb{R}

instead of three separate cases (which would still be correct). If you had found a solution to a differential equation that was

y = e^C e^{(x^3/3)}

(no absolute value) then the only possible solution would have been

y = A e^{(x^3/3)}

with

A > 0

 

[Matt1234]

Thank you amok, halls of ivy and unit.