Find the general solution of the given differential equation....

[Rijad Hadzic]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

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

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

The interval of solution according to my book is (-inf,inf)

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

 

[epenguin]

It's just that ln(0) = -∞ isn't it?

There is a solution for an infinite range of x but not y.?

 

[Rijad Hadzic]

It's just that ln(-∞) = 0 isn't it?

There is a solution for an infinite range of x but not y.?

Sorry I'm having trouble trying to understand what you're trying to get across!

Yes ln(-inf) = 0, but I don't see how that relates to this problem.

Wouldn't c/e^(x^3) be 0 at x = ln(0)^(1/3)

The book says the interval of solution is (-inf,inf)

But c/e^(x^3) is 0 at x = ln(0)^(1/3)

My logic is the interval of solution isn't (ln(0)^(1/3), inf) because ln(0)^(1/3) is not a number that you can find on the number line. Is my reasoning correct?

 

[Mark44]

Sorry I'm having trouble trying to understand what you're trying to get across!

Yes ln(-inf) = 0, but I don't see how that relates to this problem.

The "interval" is the set of x values for which the solution is defined.

Wouldn't c/e^(x^3) be 0 at x = ln(0)^(1/3)

$\frac c {e^{x^2}}$ is never zero. ln(0) is undefined.

The book says the interval of solution is (-inf,inf)

But c/e^(x^3) is 0 at x = ln(0)^(1/3)

No

My logic is the interval of solution isn't (ln(0)^(1/3), inf) because ln(0)^(1/3) is not a number that you can find on the number line. Is my reasoning correct?

No. See above.

 

[Rijad Hadzic]

Gotcha Mark. So main takeaway is that since ln(0) is undefined it cannot be in my interval of solution. e^x is cont on (-inf,inf) so the answer is (-inf,inf)

 

[epenguin]

I woke up to the thought "silly me". I stand by my first sentence bar not everyone approves this manner of speaking any more but I trust all men of good will know what I mean.

For the second the answer would be no - you can have any x, y in the plane and some solution goes through it. Moreover there is nothing unusual about your d.e. - more physical d.e.'s than not behave like it! - exponential decay towards a point that is never reached. Like dy/dx = -k²y. And just put y = ⅓ as initial value in your d.e.. you never move from where you start.(that is the only case where you get to your troublesome point!)

 

[Mark44]

Gotcha Mark. So main takeaway is that since ln(0) is undefined it cannot be in my interval of solution.

No, this isn't right. You're confusing the domain of the solution with possible output values. The log function has nothing to do with anything here.
Your DE is $y' + 3x^2y = x^2$, and your solution is $y = f(x) = \frac 1 3 + Ce^{-x^3}$
f is defined and continuous on the entire real line, and so is its derivative. The value of f(0) depends on C (i.e., $f(0) = \frac 1 3 + C$), and C in turn would be specified if an initial condition were present.

e^x is cont on (-inf,inf) so the answer is (-inf,inf)

If by "answer" you mean the interval on which the solution is defined, then yes.

 

[Ray Vickson]

It's just that ln(-∞) = 0 isn't it?

There is a solution for an infinite range of x but not y.?

I don't think that $e^0 = -\infty$, as your equation claims.

However, I could go with $\ln(0) = -\infty$ because $e^{-\infty} = 0$.

 

[epenguin]

I don't think that $e^0 = -\infty$, as your equation claims.

However, I could go with $\ln(0) = -\infty$ because $e^{-\infty} = 0$.

Oops Late night posting, is it OK now?

 

[Mark44]

However, I could go with $\ln(0) = -\infty$ because $e^{-\infty} = 0$.

I get what this is saying, but it's pretty sloppy mathematically. 0 isn't in the domain of the ln function, nor is $-\infty$ in the domain of the exponential function, considering the "ordinary" real numbers.

Better would be $\lim_{x \to 0^+}\ln(x) = -\infty$ and $\lim_{x \to -\infty}e^x = 0$.