Pre-calc related question for the interval of a solution

[Rijad Hadzic]

Homework Statement

For a differential equation I have solution

y= (1/3) + ce^(-x^3) where c is a constant

The interval of solution is (-inf,inf)

that makes sense to me, since e^x never has a value of y that equals zero.

Edit: this is the original question:

Find the general solution of the given differential equation. Give the largest interval I over which the interval is defined.Determine wether there are any transient terms in the general solution

5.

\frac {dy}{dx} + 3x^2y = x^2

e^{\int {3x^2 dx}} -->>> e^{x^3}

e^{x^3} y = \int {e^{x^3}x^2 }

y = (1/3) + ce^{-x^3}

Homework Equations

The Attempt at a Solution

It makes sense when you look at a graph of e^x... but if I set e^(x^3) = 0, and I take ln for both side, and get x = ln(0)^(1/3)

wouldn't ln(0)^(1/3) be a number, and that number make the function undefined?

I'm trying to understand why it makes sense graphically but doesn't make sense logically.

My reasoning would be as follows: there is not an x value for ln(0)^1/3.

Meaning, if you look on the number line, no value of ln(0)^1/3 exists in the domain.

This reasoning makes sense to me but I feel like I'm not grasping the whole picture... can anyone help me out by clarifying? Can anyone help me out by pointing any faulty statements out? Again I thank you guys.

 

[Rijad Hadzic]

Mod note: This post really should be in the Calculus & Beyond section, but I didn't want to make the extra effort of starting a new thread in that section.
Another question just so I don't have to make another thread guys...

say I have solution

y = -xcos(x) + cx

Why would the interval of solution be (0,inf)? Couldn't (-inf,inf) be a solution as well? Because even if x = 0, you simply get -(0)(1) + c(0) = 0, which is a real number, so why can't it be (-inf,inf)??

This one I have no answer for myself because 0 IS on the real number line..Edit: here is the original questionFind the general solution of the given differential equation. Give the largest interval I over which the interval is defined.Determine wether there are any transient terms in the general solution11.

x\frac {dy}{dx} - y = x^2sin(x)
\frac {dy}{dx} + x^{-1}y = sin(x)

e^{-\int {x^{-1} dx }} = 1/x

d/dx[yx^{-1}] = sin(x)

yx^{-1} = -cos(x) + c

y = -xcos(x) + cx

I don't understand why the interval of solution isn't (-inf, inf), as that would be the largest solution.

 

[BvU]

What about c ? does it have to be positive ?

 

[Rijad Hadzic]

What about c ? does it have to be positive ?

For my second post, right? I don't see why c would have to be positive..

 

[BvU]

You created confusion by posting two questions in one thread.
$e^x$ can take on any positive value.
$c\,e^x$ can take on any value.

$\cos x$ can take on any value in [-1,1]
$\pm x\cos x$ can take on any value
$\pm x\cos x + c\, x$ can take on any value

 

[Rijad Hadzic]

You created confusion by posting two questions in one thread.
$e^x$ can take on any positive value.
$c\,e^x$ can take on any value.

$\cos x$ can take on any value in [-1,1]
$\pm x\cos x$ can take on any value
$\pm x\cos x + c\, x$ can take on any value

Which is why I don't understand why the book make the answer for y = -xcos(x) + cx having a solution of (0,inf)

Why would it not be [0, inf]?

Going one step further, why not (-inf,inf), wouldn't that be the largest possible interval?

 

[BvU]

Perhaps you can copy the exact and complete problem statement ?
(That's why the template in the homework is so useful...)

 

[BvU]

Other question: what do you mean with

The interval of solution is (-inf,inf)

since we have terms like range and domain to distinguish the intervals for independent variable and function value

 

[Rijad Hadzic]

Other question: what do you mean with
since we have terms like range and domain to distinguish the intervals for independent variable and function value

Edited the OP with the first question.

About to do the second question as well.

I thought Interval of solution only applies to domain? Meaning whatever value you have for your interval of solution it will be a solution to the differential equation.Updated post #2 with the original question that was asked

 

[Mark44]

Rijad, please don't post two different questions in one thread -- start a new thread for each.

 

[Rijad Hadzic]

ok my bad Mark I just didnt want to congest the fourm, and they were both questions of the same nature but Ill fix it next time.

I saw your mod note. Ill make two threads in the calculus section and tidy it up a bit and delete this one, the least I can do for the help provided..

 

[Mark44]

Thread closed at OP's request.