# Differential Equation - Disappearing Term

1. Oct 16, 2015

### BOAS

Hi,

I am struggling to find the solution to the following equation. I can't account for the exponential term, so clearly something is going wrong...

1. The problem statement, all variables and given/known data

Find the general solution to $x' = tx + 6te^{-t^2}$ where $x(t)$.

2. Relevant equations

3. The attempt at a solution

Consider $x' - tx = 0$

$\frac{dx}{dt} = tx$

$\frac{dx}{t} = t dt$
Mod note: the above should be $\frac{dx}{x} = t dt$
integrating and exponentiating gives $x = Ae^{\int t dt}$. Let $I = \int t dt$

$x = A e^{I}$ and rearranging gives $A = x e^{-I}$

Differentiating yields $A' = x'e^{-I} - t x e^{-I}$

Factoring out the exponential term, and recognising that this is the same as the initial equation.

$A' = e^{-I}(x' - tx) = e^{-I}(6te^{-t^{2}})$

$A = \int e^{-I}(x' - tx) dt = \int e^{-I}(6te^{-t^{2}}) dt$ which simplifies to

$A = \int e^{-\frac{3}{2} t^{2} + c} 6t dt$

Using the substitution $u = - \frac{3}{2} t^{2}$ I find that

$A = - 2 \int e^{u} du = -2e^c e^{- \frac{3}{2} t^{2}} + c$

$x = e^I A = e^{- \frac{1}{2} t^2}(-2 e^{- \frac{3}{2} t^{2}} + c)$

This is not a solution to my differential equation. The $6t e^{-t^2}$ term is nowhere to be seen.

What am I doing wrong?

Last edited by a moderator: Oct 16, 2015
2. Oct 16, 2015

### epenguin

I found easier than going through your arguments, just noticing that differentiating e-t2 gives you something very suggestive for a solution which I got.

3. Oct 16, 2015

### geoffrey159

The steps to solve a first order differential equation of the type $y' + a y = b$, where $a,b$ are continuous, real valued functions, are
1 - Find the general solution $y_0$ of the equation when $b=0$, these are $\{ \lambda e^{-A}, \lambda\in\mathbb{R}, A \text{ primitive of } a \}$
2 - Find a particular solution $y_1$ to the equation. In your case it requires no calculations.
3 - Add these solutions to have the general solution of the equation

4. Oct 16, 2015

### BOAS

Hmm, you are right.

$x = \frac{1}{2} e^{t^{2}} - 3 e^{-t^{2}} + C$

Last edited: Oct 16, 2015
5. Oct 16, 2015

### geoffrey159

Your answer is incomplete because the solution is not unique unless you have a condition such as $x(t_0) = y_0$

6. Oct 16, 2015

### geoffrey159

Still incomplete, even with $+C$ . Read post #3

7. Oct 16, 2015

### Ray Vickson

Your main error is that initially, $x$ was a solution of $x' - tx = 0$. Then you put $x' - tx = 6 t e^{-t^2}$ in your expression for $A'$. That is incorrect: as originally defined, $A$ was a constant, with $A' = 0$, so you cannot just pull a switcheroo and suddenly make $A'$ non-zero, without at least changing the meaning of $A$. Go back to tried-and-true methods, as presented in your textbook or (probably) your course notes.

8. Oct 16, 2015

### epenguin

I don't think that's quite right, it seems to me

x = -2 e-t2 + C

will do it.

Last edited: Oct 16, 2015
9. Oct 16, 2015

### geoffrey159

No, the solutions are the $x_\lambda(t) = -2e^{-t^2} + \lambda e^\frac{t^2}{2}$, $\lambda\in\mathbb{R}$

10. Oct 16, 2015

### epenguin

I only said #2 mine was A solution.

11. Oct 17, 2015

### BOAS

The method I used is the same (I think) as presented in 'Maths Methods For the Physical Sciences' - Mary Boas, on page 346

It claims that $y = e^{-I} \int Q e^{I} dx + C e^{-I}$ is the general solution to a linear first order differential equation of the form $y' + Py = Q$

12. Oct 17, 2015

### BOAS

I appreciate that you have all taken the time to help me here, but I think it's best I go back to square one, as you have highlighted some pretty core misunderstandings.

13. Oct 17, 2015

### vela

Staff Emeritus
The point here is that if you multiply $x' - tx$ by $e^{-I}$, you turn it into the derivative of $x e^{-I}$.

This line would be okay if you didn't say it was equal to A', which is equal to 0 because A is a constant.

You made a sign error. $I$, as you defined it above, is equal to $+\frac{t^2}{2}$, which gives you $x(t) = -2 e^{-t^2} + c e^{t^2/2}$, which is the correct solution.

Note that for the general case you cited above, $I$ is defined as $\int p(x)\,dx$, but the way you did it in this problem, you set $I = -\int p(x)\,dx$. That's why you have the difference in sign.

14. Oct 17, 2015

### geoffrey159

The thing is that you need to be very careful with algebraic manipulations.
For exemple, a division of your original DE by a function that has got one or several zeros in the interval you are interested in will force you to solve your DE in several sub-intervals and reconnect by continuity and differentiability the solutions found at the bounds.
If you can't do these connections, your solution set is $\emptyset$ since your $y$ (or $y'$) is not globally continuous.
But you will never see that if you just divide and integrate.

15. Oct 17, 2015

### BOAS

I have worked through the problem again and arrived at the correct solution.

What I don't understand is why the derivative of the solution is not $= tx + 6t e^{-t^{2}}$, but rather just $tx$

16. Oct 17, 2015

### vela

Staff Emeritus
You must be making a mistake when differentiating. I find $x'(t) = 4 e^{-t^2} + c t e^{t^2/2}$. If you subtract $xt$ from that, you're left with $6te^{-t^2}$ as expected.

17. Oct 17, 2015

### BOAS

You are correct.

Thank you.

18. Oct 17, 2015

### Ray Vickson

I hope she does not actually say that, because as you have written it, it is either (i) wrong; (ii) inaccurate; (iii) misleading, or (iv) error-prone. What IS true is that the solution of $y/ + P y = Q$ has the form
$$y =c e^{-q(t)} + e^{-q(t)} \int_0^t e^{q(s)} Q(s) \, ds = c e^{-q(t)} + \int_0^t e^{q(s) - q(t)} Q(s) \, ds,$$
where $q(t) = \int_0^t Q(s) \, ds$. The correct formula involves specifying function arguments $q(t)$, plus distinguishing between the variable $t$ in the solution $y(t)$ and the integration dummy variable $s$. Just writing $y = e^{-I} \int Q e^{I} dx + C e^{-I}$ leaves way too much to the imagination and leaves way too much room for mistakes, etc.

19. Oct 18, 2015

### BOAS

My mistakes aside, my argument follows fairly closely the steps outlined in the book (see attached).

Could you recommend a more rigorous text? I am concerned that many of the explanations will be of similar calibre.

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20. Oct 18, 2015

### geoffrey159

It is incorrectly explained.
Just an exemple, when she says around (3.3) second scan "We obtain a solution as follows : $\frac{dy}{y} = -P dx$ ...... $\ln(y) = ...$ ".

She doesn't know that $y$ does not cancel at some point on the interval of interest. You know it a posteriori, once you have found $y$.
So she assumes that the solution has no zero without even knowing it, and she uses divisions by $y$ and uses logarithm of $y$ at all points. It is incorrect !