I'm trying to learn DE independently, so if this is insanely stupid,

[TylerH]

I'm trying to learn DE independently, so if this is insanely stupid, try not to make me cry. :)

\frac{dy^2}{dx^2}+d\frac{dy}{dx}=x
dy^2+dxdy=xdx^2
(y + C)dy+(x+D)dy=(\frac{x^2}{2}+E)dx This is the line that gives me the incling I'm way off... this could be what is below, or could simplify the left to d((x+c)(y+d))
\frac{y^2}{2}+Cy+\frac{x^2}{2}+Dx+E=\frac{x^3}{6}+Fx
and so on...

If this is wrong, what is the correct method?

 

[Mark44]

I'm trying to learn DE independently, so if this is insanely stupid, try not to make me cry. :)

I hate to break it to you, but you're way off.

\frac{dy^2}{dx^2}+d\frac{dy}{dx}=x
dy^2+dxdy=xdx^2
(y + C)dy+(x+D)dy=(\frac{x^2}{2}+E)dx This is the line that gives me the incling I'm way off... this could be what is below, or could simplify the left to d((x+c)(y+d))
\frac{y^2}{2}+Cy+\frac{x^2}{2}+Dx+E=\frac{x^3}{6}+Fx
and so on...

If this is wrong, what is the correct method?

This is an example of a nonhomogeneous (essentially, the right side isn't zero), linear, constant coefficient differential equation.

First, start with the homogeneous equation, which is y'' + y' = 0. In your book, look at the section where they solve homogeneous 2nd order DEs. They'll probably talk about characteristic equations, which for this DE is r² + r = 0. The left side factors to r(r + 1) = 0, so r = 0 or r = -1.

This means that two linearly independent solutions to the homogeneous DE are y = e^0t and y = e^-1x, or more simply, y = 1 and y = e^-x.

All solutions of the homogenous DE consist of all linear combinations of these two functions. So the general solution of the homogeneous DE is y = c₁ + c₂e^-x.

Now you try to find a particular solution of the nonhomogeneous equation. In one approach, you make educated guesses about what is likely to work. There's another approach, but the explanation is a bit lengthy.

For this particular problem, a particular solution is y_p = Ax + Bx². By substituting this equation into the DE you can solve for the constants A and B.

The general solution to the nonhomogeneous problem is y = y = c₁ + c₂e^-x + y_p.

 

[AlephZero]

I'm trying to learn DE independently, so if this is insanely stupid, try not to make me cry. :)

\frac{dy^2}{dx^2}+d\frac{dy}{dx}=x
dy^2+dxdy=xdx^2

That is horribly wrong. It's hard diagnose how to put you on the right track from just one mistake as "big" as this, but you may have got confused by the notation. It should be

\frac{d^2y}{dx^2}

not your

\frac{dy^2}{dx^2}

and it doesn't mean the same as

\frac{(dy)^2}{(dx)^2}

If you did want to go down that route, you would start with something like

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

\left(\frac{d}{dx} + 1 \right) \left( \frac{d}{dx} \right) y =x

But this isn't the simplest way to go - see post #2.

 

[TylerH]

So differentials can't be treated like they are multiplied? ie If you can make it \left(\frac{d}{dx}+1\right)(\left \frac{d}{dx} \right)y=x, it doesn't mean you can make it \left(\frac{d}{dx}+1\right)(\left d \right)y=xdx?

Wow... I'm SOO confused... As you, Aleph, said, it's probably a combination of the fact that differentiation and integration having similar semantics to multiplication, both distribute over addition (or, at least, that's the way I took it) and one is similar in notation.

 

[Mark44]

So differentials can't be treated like they are multiplied? ie If you can make it \left(\frac{d}{dx}+1\right)(\left \frac{d}{dx} \right)y=x, it doesn't mean you can make it \left(\frac{d}{dx}+1\right)(\left d \right)y=xdx?

I suppose it makes sense, but it's not of any use.

The reason for writing
\left(\frac{d}{dx}+1\right)(\left \frac{d}{dx} \right)y=x
as AlephZero has it, or
(D + 1)(D)y = x, as I wrote it, is the similarity to the characteristic equation. If that's not something you've learned about yet, then all this is probably over your head.

Wow... I'm SOO confused... As you, Aleph, said, it's probably a combination of the fact that differentiation and integration having similar semantics to multiplication, both distribute over addition (or, at least, that's the way I took it) and one is similar in notation.

 

[coelho]

Well... actually you could solve

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

by writing it as

\frac{d}{dx}\left(\frac{dy}{dx}+y\right)=x

and integrating in dx both sides, to get:

\int \frac{d}{dx}\left(\frac{dy}{dx}+y\right) dx = \int x dx
\frac{dy}{dx}+y = \frac{x^2}{2}+c

then solving this simpler (1st-order) equation by whatever methods you wish.

BUT this does not mean that differentials can be treated as if they were simply being multiplied. There is a bit more of math behind that.

 

[AlephZero]

You are right, you can often do thiings with dx and dy that look like multiplying fractions, when they are written down, but you need to understand what the notation means, otherwise you are just shuffling symbols around on a piece of paper, and the result may be wrong, or even meaningless.

It might help to think of "d/dx" not as a fraction, but as a function, the same as "sin" or "log". "sin y" means "do something to y to produce a value". Similarly "dy/dx" means "do something to (i.e. differentiate) y to produce its differential".

And "d^2y/dx^2" means "apply the d/dx operation to (i.e differentiate) dy/dx"

Now, think back to how we define what "dy/dx" means. It means "the limit of \Delta y / \Delta x as \Delta x tends to 0".

When you write

dy/dx = x, therefore dy = x dx

That is also a statement about limits, not a fraction. It means "the limit of \Delta y = the limit of x \Delta x, as \Delta x tends to 0".

The nice thing about the "fraction" notation is that when it means something, the symbols really do behave the same as fractions. But you have to be careful not to write down things that are meaningless. For eaxmple you wouldn't (I hope!) write something like

y = sin(x), therefore sin = y/x
or even sillier,
s = y / in(x).

Because you (apparently) haven't understood what the differental equation notation really means, you are trying to write things that are just as wrong as those examples.

If you tell us what courses you have already done, and what books or websites you are using to study this, and you will probably get some good advice on whether those are good sources or not.

Hope this helps

 

[TylerH]

I've taken calc I, but I know most of calc II from self learning (Yes... you may want to question this, seeing the self learning facilitated debacle I've made here, but trust me.). I learned linear separable diff eqs from a friend and ones solvable by e^∫p(x)dx from Wikipedia.

My understanding is that "d" is the operator/function and dx is what is produced when you differentiate a function in x. The /dx is just dividing off the dx produced, right?

ie f(x)=x^2
df=2xdx
df/dx=f'(x)=2x

 

[AlephZero]

OK, this may be one of the dangers of learning from Wikipediia. Unless you know what the Wiki page is about before you read it, you can't tell if it is talking about an simple introduction to the subject, or an advanced modern math version. Often one page is a mixture of both.

IMO, to get hold of the basic ideas of calculus the best place to start is a geometrical interpretation of "dy/dx" (as a single symbol) as the slope of a tangent to the graph of the function that defines y against x, and the idea that you find the tangent by drawing a line through two points on the curve close together, and taking the limit as the points get closer together.

Then, you can use the "epsilon-delta" definition of what a limit means, both to prove the standard results about how to differentiate functions, the product and chain rules, etc, and see how "differentials" like dy and dx relate to the limit definition. If you go that way, you don't need the idea of "d" on its own meaning anything - not until you get to grad school after a math major, at least.

You would probably be better spending few dollars on a high-school textbook than relying only on web sites. If you want to learn from the web, there have been some good recommendations for the Khan academy videos.
https://www.physicsforums.com/showthread.php?t=354091

 

[Studiot]

My understanding is that "d" is the operator/function and dx is what is produced when you differentiate a function in x. The /dx is just dividing off the dx produced, right?

There is another approach called the D (note the capital) operator method.

I would not recommend branching off to this at this time, until you have properly learned the meaning of

\frac{{dy}}{{dx}}

You already have several solution methods proposed in this thread so adding more will, I think, cause you confusion.

 

[TylerH]

I've already done all that! Honestly, I appreciate the gentle intro, but I've done all of it. Where I was messing up was that I was thinking d²y=dy². Now I see the first means to differentiate y twice, whereas the second means the y was differentiated twice(or something like that, see below).

As for the meaning of dy/dx: dy is the infinitesimal change in y and dx is the infinitesimal change in x, you put them in ratio and you get the finite result of how much they change in relation to each other. Right?

My philosophy is to understand the underlying mechanics first and learn methods that use them later. What you seem to be getting at. I'm finding it hard to understand what exactly is means to have a differential after an expression ie "2xdx." I realize this is a deep question. There is no need to answer it. I'm sure it is explained somewhere, I just don't know what to search. Just some keywords would suffice.

Thanks for your patience. :)

 

[Studiot]

Using the \frac{{dy}}{{dx}} notation (that's all it is) has the great advantage that it does the book- keeping for you.

It is fashionable, these days, not to complete the mathematical sentences, which is OK when you are slick and quick at the stuff, but can also lead to confusion when learning.

Take

y = 3x² + 2.

\frac{{dy}}{{dx}} = 6x.

My old maths teacher would have marked that wrong unless I stated "d.w.r.x." (differentiate with respect to x)

Writing it like this is easy to see how you get

dy = 6xdx

and

\int {dy = \int {6xdx} }

Note that for integration the dx - or dy or dsomething - is necessary. Again folks often omit it when talking about integrals in general but you have to add it back in when actually working one out.

Incidentally, the notation y' and y'' is usually reserved for differentiation with respect to time.

 

[Mark44]

Using the \frac{{dy}}{{dx}} notation (that's all it is) has the great advantage that it does the book- keeping for you.

It is fashionable, these days, not to complete the mathematical sentences, which is OK when you are slick and quick at the stuff, but can also lead to confusion when learning.

Take

y = 3x² + 2.

\frac{{dy}}{{dx}} = 6x.

My old maths teacher would have marked that wrong unless I stated "d.w.r.x." (differentiate with respect to x)

That seems unnecessarily pedantic. For y as defined, the symbol dy/dx contains information that y was differentiated with respect to x. Given the definition for y, I don't see anything wrong with saying dy/dx = 6x.

Writing it like this is easy to see how you get

dy = 6xdx

and

\int {dy = \int {6xdx} }

Note that for integration the dx - or dy or dsomething - is necessary. Again folks often omit it when talking about integrals in general but you have to add it back in when actually working one out.

Incidentally, the notation y' and y'' is usually reserved for differentiation with respect to time.

That's not my experience. Newton's dot notation refers to differentiation w.r.t. time.

I.e.,
\dot{y} = \frac{dy}{dt}
\ddot{y} = \frac{d^2 y}{dt^2}

Is that what you meant?

 

[Studiot]

That seems unnecessarily pedantic.

The basic principle was that every line you wrote had a reason or justification appended.
That also aided marking schemes where separate marks were allocated for method and result.
It is interesting to note the exhortation in every computer programming text to annotate and lay out computer programs in this way for ease of subsequent understanding.

Is that what you meant?

Yes that is also used.

 

[Mark44]

The basic principle was that every line you wrote had a reason or justification appended.
That also aided marking schemes where separate marks were allocated for method and result.

OK, that sounds more reasonable. My grade 10 high school geometry class was like this, where you wrote a statement, and then gave justification for it. I don't recall subsequent classes requiring the same level of justification.

It is interesting to note the exhortation in every computer programming text to annotate and lay out computer programs in this way for ease of subsequent understanding.



Yes that is also used.

As I understand it, Newton used dot notation exclusively, and to him all differentiation was with respect to time t. I don't know who came up with the prime and double-prime notation to represent first and second derivatives. The dot notation seems to be used mostly in physics classes, at least here in the US. Math classes and textbooks here use the primes notation, with the meaning of y' given in the context in which it appears or explicitly in notation such as y'(x). It might be different in GB, from whence you hail, Studiot.

 

[TylerH]

I don't know who came up with the prime and double-prime notation to represent first and second derivatives.

http://en.wikipedia.org/wiki/Derivative#Lagrange.27s_notation

 

[Mark44]

I don't know who came up with the prime and double-prime notation to represent first and second derivatives.

http://en.wikipedia.org/wiki/Derivative#Lagrange.27s_notation

Now I know! Thanks, Tyler.

 

[TylerH]

After doing a little research on him, I have a theory on why his notation is used so prevalently in diff eq. He succeeded Euler, who discovered characteristic equations and did lots of other founding work on DE (Euler method of numerical solving comes to mind), at the Prussian Academy of Sciences. One can guess that, with all the discoveries being made by Euler at the time, the Prussian Academy of Sciences probably became the premier DE school. It also probably retained it's status when Lagrange took over. He probably pushed the usage of his notation.

Not that I'm saying all he did was piggy back off the immense awesomeness of Euler. Apparently, he did some serious number theory stuff.

Now, to prove I'm no dumby head: I think I've figured it out:
y''+y'=e^x
y=e^rx+f(x) You called f(x) "y_p"
r²+r=0
r={-1,0}
y=C+De^-x+f(x)
f(x)=Ae^x Not really sure how to justify this. It just came to me.
(C+De^-x+Ae^x)''+(C+De^-x+Ae^x)'=e^x
solves to
A=1/2
y=C+De^-x+.5e^x

EDIT: Should I leave C and D or let them be 0?

Is there a name for the nonguess and check technique used for finding f(x)/y_p?

 

[Mark44]

After doing a little research on him, I have a theory on why his notation is used so prevalently in diff eq. He succeeded Euler, who discovered characteristic equations and did lots of other founding work on DE (Euler method of numerical solving comes to mind), at the Prussian Academy of Sciences. One can guess that, with all the discoveries being made by Euler at the time, the Prussian Academy of Sciences probably became the premier DE school. It also probably retained it's status when Lagrange took over. He probably pushed the usage of his notation.

Not that I'm saying all he did was piggy back off the immense awesomeness of Euler. Apparently, he did some serious number theory stuff.

Now, to prove I'm no dumby head: I think I've figured it out:
y''+y'=e^x
y=e^rx+f(x) You called f(x) "y_p"
r²+r=0
r={-1,0}
y=C+De^-x+f(x)
f(x)=Ae^x Not really sure how to justify this. It just came to me.
(C+De^-x+Ae^x)''+(C+De^-x+Ae^x)'=e^x
solves to
A=1/2
y=C+De^-x+.5e^x

EDIT: Should I leave C and D or let them be 0?

Is there a name for the nonguess and check technique used for finding f(x)/y_p?

This won't work, and you can confirm that it doesn't work by substituting your f(x) into the nonhomogeneous DE.

The complementary solution y_c, is the general solution to the homogeneous equation. In this case, y_c = c₁ + c₂e^-x.

What I called the particular solution y_p (and you're calling f(x)) is a solution to the nonhomogenous DE, y'' + y' = x.

You have f(x) = Ae^x. To see that it doesn't work, note that f(x) = f'(x) = f''(x) for this function, so we have Ae^x + Ae^x = x, or 2Ae^x = x. This has to be an identity - true for all x - and that isn't the case here.

For the equation y'' + y' = x, a particular solution is y_p = f(x) = Ax + Bx².

As for the name of the method to find y_p, I'm not sure there's a name. What I did was to turn the 2nd order nonhomogeneous equation y'' + y' = x into a 4th order homogeneous equation, y⁽⁴⁾ + y⁽³⁾ = 0, using what is called the method of annihilators.

As mentioned earlier, the original DE can be represented as D(D + 1)y = x. By noting that the D² operator (i.e., the second derivative with respect to x) produces 0 when applied to x, I can tack on a "factor" of D² to both sides of the original DE.

This gives D²D(D + 1)y = D²(x) = 0, which is now a homogeneous equation.

The characteristic equation is r³(r + 1) = 0, and the roots are r = 0 (three times) and r = -1.

This gives e^0x and e^-1x as solutions, but I need four independent solutions, not just two. The trick is that I can get two more functions for my set by tacking factors of x and x² onto the function that's associated with the repeated roots.

This gives me {e⁰, xe⁰, x²e⁰, e^-x} for my set of functions, or more simply, {1, x, x², e^-x}.

Any solutions of the fourth order homogeneous DE will be a linear combination (sum of constant multiples of) these four functions.

Going back to the original problem, two of these solutions (1 and e^-x) are solutions to the homogeneous problem y'' + y' = 0. The other two (x and x²) are what I picked for my "guess," looking at all linear combinations, of course. In short, y_p = Ax + Bx².

 

[TylerH]

LOL @ misunderstanding! I changed the problem so it was obvious I wasn't just copying you. That's why the first equation of the many in my last post was y''+y'=e^x. I hope you're not meaning (.5e^x)''+(.5e^x)'=.5e^x+.5e^x≠e^x. :)

EDIT: I forgot: Thanks, again, for your patience.

 

[TylerH]

This gives e^0x and e^-1x as solutions, but I need four independent solutions, not just two. The trick is that I can get two more functions for my set by tacking factors of x and x² onto the function that's associated with the repeated roots.

This gives me {e⁰, xe⁰, x²e⁰, e^-x} for my set of functions, or more simply, {1, x, x², e^-x}.

How did know you choose x and x²? Heuristics is a fine answer if there is no other explanation.

Learning integration has taught me that heuristics, unfortunately, do have a place in math. (Referring to substitution.) Destroying my view of math as the only sanctuary of determinism, this made me sad, at the time. jk :P

Great explanation, by the way.

 

[Mark44]

LOL @ misunderstanding! I changed the problem so it was obvious I wasn't just copying you. That's why the first equation of the many in my last post was y''+y'=e^x. I hope you're not meaning (.5e^x)''+(.5e^x)'=.5e^x+.5e^x≠e^x. :)

EDIT: I forgot: Thanks, again, for your patience.

I didn't realize that you had changed the right side on the DE.

 

[Mark44]

How did know you choose x and x²? Heuristics is a fine answer if there is no other explanation.

I'm not sure what the origin of this is. There's a lot of mathematics that has arisen because someone discovered a useful technique.

Learning integration has taught me that heuristics, unfortunately, do have a place in math. (Referring to substitution.) Destroying my view of math as the only sanctuary of determinism, this made me sad, at the time. jk :P

Great explanation, by the way.

 

[Studiot]

Looking over a selectioon of my books, you are right Mark, the dot notaion is more used for time than the prime, which is used for many other things as well.
It is, perhaps, not suprising that the Europeans should introduce the prime notation since most of their alphabets contain accented letters anyway.

On another note, Tyler, did you understand my comment about integrals which was directly addressed to one of your queries/difficulties?

I'm finding it hard to understand what exactly is means to have a differential after an expression ie "2xdx."

I am trying to supplement, not supplant, the excellent job Mark is doing at explanation here, but I agree with AZ that you have some of the basics muddled up.

Continuing this make sure you see the difference between

\frac{{{d^2}y}}{{d{x^2}}}\quad and\quad {\left( {\frac{{dy}}{{dx}}} \right)^2}

In prime terms this is the difference between y'' and (y')²

go well

 

[TylerH]

On another note, Tyler, did you understand my comment about integrals which was directly addressed to one of your queries/difficulties?



I am trying to supplement, not supplant, the excellent job Mark is doing at explanation here, but I agree with AZ that you have some of the basics muddled up.

Continuing this make sure you see the difference between

\frac{{{d^2}y}}{{d{x^2}}}\quad and\quad {\left( {\frac{{dy}}{{dx}}} \right)^2}

In prime terms this is the difference between y'' and (y')²

go well

Yes, I believe I understood. The thanks were directed at all, not just Mark. :)

I still don't know what it means to have a differential next to an expression, though... (ie what the implied function is.)

 

[TylerH]

I'm not sure what the origin of this is. There's a lot of mathematics that has arisen because someone discovered a useful technique.

I did some research. http://en.wikipedia.org/wiki/Characteristic_equation#Repeated_real_roots

For n repeated roots of the characteristic equation at x₀: y=e^x₀x(A+Bx+...+Zx^n-1).