Pre-calc related question for the interval of a solution

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In summary, the general solution of the given differential equation is y=-xcos(x)+c. There are no transient terms in the general solution.
  • #1
Rijad Hadzic
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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.

[itex] \frac {dy}{dx} + 3x^2y = x^2 [/itex]

[itex] e^{\int {3x^2 dx}} [/itex] -->>> [itex] e^{x^3} [/itex]

[itex] e^{x^3} y = \int {e^{x^3}x^2 } [/itex]

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

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.
 
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  • #2
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.

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

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

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

[itex] yx^{-1} = -cos(x) + c [/itex]

[itex] y = -xcos(x) + cx [/itex]

I don't understand why the interval of solution isn't (-inf, inf), as that would be the largest solution.
 
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  • #3
What about c ? does it have to be positive ?
 
  • #4
BvU said:
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..
 
  • #5
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
 
  • #6
BvU said:
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?
 
  • #7
Perhaps you can copy the exact and complete problem statement ?
(That's why the template in the homework is so useful...)
 
  • #8
Other question: what do you mean with
Rijad Hadzic said:
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
 
  • #9
BvU said:
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
 
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  • #10
Rijad, please don't post two different questions in one thread -- start a new thread for each.
 
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  • #11
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..
 
  • #12
Thread closed at OP's request.
 

FAQ: Pre-calc related question for the interval of a solution

1. What is an interval of a solution in pre-calculus?

An interval of a solution in pre-calculus is a range of values for a variable that satisfies the given equation or inequality. It is represented by a set of numbers between a lower and upper bound.

2. How do you determine the interval of a solution in pre-calculus?

To determine the interval of a solution, you first need to solve the given equation or inequality. Then, you need to identify the values of the variable that satisfy the equation or inequality. These values will form the lower and upper bound of the interval of solution.

3. Can there be multiple intervals of solution for one pre-calculus problem?

Yes, there can be multiple intervals of solution for one pre-calculus problem. This can happen when the given equation or inequality has multiple solutions that fall within different ranges of values for the variable.

4. How do you represent the interval of a solution in pre-calculus?

The interval of a solution in pre-calculus is typically represented using interval notation, which uses brackets or parentheses to indicate whether the endpoints are included or excluded. For example, [a,b) means the interval includes the lower bound a but not the upper bound b.

5. Why is it important to find the interval of a solution in pre-calculus?

Finding the interval of a solution is important in pre-calculus because it helps us understand the possible values of the variable that will satisfy the given equation or inequality. This information can be used to make decisions in real-world applications and to better understand the behavior of functions.

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