Differential Equation - Disappearing Term

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

Homework Equations

The Attempt at a Solution

[/B]
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?

 

[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.

 

[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

 

[BOAS]

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.

Hmm, you are right.

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

 

[geoffrey159]

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

 

[geoffrey159]

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

 

[Ray Vickson]

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

Homework Equations

The Attempt at a Solution

[/B]
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?

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.

 

[epenguin]

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

x = -2 e^-t2 + C

will do it.

 

[geoffrey159]

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

x = 2 e^-t2 + C

will do it.

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

 

[epenguin]

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

I only said #2 mine was A solution.

 

[BOAS]

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.

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 $y' + Py = Q$

 

[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.

Thank you for your help.

 

[vela]

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

Homework Equations

The Attempt at a Solution

[/B]
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 $A' = x'e^{-I} - t x e^{-I}$

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

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

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

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.

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

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.

 

[geoffrey159]

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.

Thank you for your help.

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.

 

[BOAS]

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.

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 $tx$

 

[vela]

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 $tx$

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.

 

[BOAS]

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.

You are correct.

I missed the minus sign in my derivative.

Thank you.

 

[Ray Vickson]

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 $y' + Py = Q$

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.

 

[BOAS]

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.

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.

 

[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 !

 

[vela]

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

I think the explanation in the text is fine. Sure, you can get really pedantic and point out that the x in the integral isn't the same as the x outside the integral, or when dividing by y, we must assume y doesn't vanish on the interval of interest, etc., but as one of my professors used to say, one should focus on the donut and not the hole.

 

[geoffrey159]

I think the explanation in the text is fine. Sure, you can get really pedantic and point out that the x in the integral isn't the same as the x outside the integral, or when dividing by y, we must assume y doesn't vanish on the interval of interest, etc., but as one of my professors used to say, one should focus on the donut and not the hole.

Why is it pedantic to say that ?

 

[vela]

Because by the time most people are learning about how to solve differential equations, they're well acquainted with the abuse of notation. To keep pointing it out over and over and over is pedantic.

 

[geoffrey159]

Whether what you think is justified or not, I'm not going to teach you that telling directly to people they are pedantic is unfriendly if not rude, am I ?
It iches me to reply but I don't want to spoil this thread with that kind of stuff.

 

[vela]

Please don't take things so personally. I'm not saying anyone is being pedantic. My point is that the author is to explain the reasoning behind where the integrating factor comes from rather than just pulling it out of thin air. Could she have devoted more space to the technical details? Sure, but doing that will tend to clutter up the presentation and detract from what she's trying to accomplish.

I realize this probably makes mathematicians cringe, but the book is, after all, intended for physics students.

 

[geoffrey159]

The notation $dx$ is an abuse of notation that mixes up a function with its image (i.e the identity function ). It makes sense with an explanation. The proof in your book is not an abuse of notation, it is wrong.
The following demo is rigorous, also easier to read in my opinion, and is based on the idea that continuous functions on an interval have primitives that are themselves continuous.

If you assume that a is continuous from $I$ into $\mathbb{R}$, then there are primitives of $a$ on $I$. Let $A$ be one of them :
$\begin{align*} y' + ay = 0 &\iff (y' + ay) e^A = 0 \\ &\iff (ye^A)' = 0 \\ &\iff (\exists c \in\mathbb{R} \ ye^A = c) \\ &\iff (\exists c \in\mathbb{R}\ y =c e^{-A} ) \end{align*}$

If you assume $b$ continuous from $I$ into $\mathbb{R}$, then $be^A$ is continuous and has primitives on $I$, $B$ for exemple. So
$\begin{align*}y' + ay = b &\iff (y' + ay) e^A = be^A \\ & \iff (ye^A)' = be^A \\ & \iff (\exists c \in\mathbb{R} \ ye^A = B +c) \\ & \iff (\exists c \in\mathbb{R}\ y = B e^{-A} + c e^{-A} )\end{align*}$

 

[Ray Vickson]

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.

That textbook presentation is inexcusably sloppy---so much so as to be virtually useless.

Please don't take things so personally. I'm not saying anyone is being pedantic. My point is that the author is to explain the reasoning behind where the integrating factor comes from rather than just pulling it out of thin air. Could she have devoted more space to the technical details? Sure, but doing that will tend to clutter up the presentation and detract from what she's trying to accomplish.

I realize this probably makes mathematicians cringe, but the book is, after all, intended for physics students.

Then I think it should be more careful, so as to not confuse the student for no really good reason. Back in the Stone Age when I was taking undergrad courses is "mathematical methods for physics", we of course, had lots of such material, and many times abuse of notation was performed/permitted. However, sometimes it was NOT, and I think this is one of those times. Go ahead and abuse notation if no serious confusion can arise, but avoid it like the plague if it can lead to uncertainty and avoidable errors.

 

[epenguin]

The notation $dx$ is an abuse of notation that mixes up a function with its image (i.e the identity function ). It makes sense with an explanation. The proof in your book is not an abuse of notation, it is wrong.
The following demo is rigorous, also easier to read in my opinion, and is based on the idea that continuous functions on an interval have primitives that are themselves continuous.

If you assume that a is continuous from $I$ into $\mathbb{R}$, then there are primitives of $a$ on $I$. Let $A$ be one of them :
$\begin{align*} y' + ay = 0 &\iff (y' + ay) e^A = 0 \\ &\iff (ye^A)' = 0 \\ &\iff (\exists c \in\mathbb{R} \ ye^A = c) \\ &\iff (\exists c \in\mathbb{R}\ y =c e^{-A} ) \end{align*}$

If you assume $b$ continuous from $I$ into $\mathbb{R}$, then $be^A$ is continuous and has primitives on $I$, $B$ for exemple. So
$\begin{align*}y' + ay = b &\iff (y' + ay) e^A = be^A \\ & \iff (ye^A)' = be^A \\ & \iff (\exists c \in\mathbb{R} \ ye^A = B +c) \\ & \iff (\exists c \in\mathbb{R}\ y = B e^{-A} + c e^{-A} )\end{align*}$

You are joking?

Please tell us you are joking!