Help me understand initial value problems!

[Budget_Player_Cadet]

So one of the exercises in my math class that I’m supposed to solve goes something along the lines of this:

"Solve the initial value problem with the differential equation of

y’(x) = sqrt(abs(y(x)))

with y(0) != 0. Solve the initial value problem with y(0)=0. What can you say about the uniqueness of the solution?"

(Not sure if “uniqueness” is the right word; the german term is “eindeutigkeit”; implying that there’s one precise solution or something like that).

So I’ve followed the typical path (reduce it to dy/sqrt(y) = dx or dy/sqrt(-y) = dx depending on whether y is positive or not, integrate) and gotten to y = 1/4(x+c)^2 (I used a substitution of -y = z for y < 0, I think that’s the way to go on that…). c is a constant from integration.

But now what? I don’t know what they mean by “solve”. I don’t know what I’m solving for. Is it just y? Is it y with a specific c? And what do they mean by “eindeutig”? In the given solution, they get as far as I got and continue with:

“y = 1/4(x+2*sqrt(y_0)^2 solves the initial value problem on y(0) != 0”

…But wouldn’t any c != 0 solve for y(0) != 0?

But where it really loses me is the second part. Solve for y(0) = 0.

The solution says that y(x) = 0 is a solution, because y’ = 0 = sqrt(0), but what does that even mean? How do you get from y(x) = (x+c)^2 * 1/4 to y(x) = 0? c = x? Is that even possible? But the next line really puzzles me:

y(x) =
{1/4(x-a)^2 for x>= a, a >= 0 }
{0 for x <= 0. }

This was the solution the professor offered. I have no** idea what a is supposed to be (well, to be fair, it’s alpha, maybe that helps someone).

This is gonna be a big part of the upcoming makeup test. Any help would be much appreciated. I just don’t understand what, exactly I’m solving for, and I have no idea what the “uniqueness quantification” is or how to demonstrate it.

[ThisUsernameIsForbidden]

So hard to read. Is there a board that allows you to embed latex?

[Budget_Player_Cadet]

I’m not a member at any and I don’t know how to use LaTeX. Probably something I should learn, I’m gonna need it next year for my bachelorarbeit anyways. I could try to simplify it… My main issue is still “what am I solving for”.

[Asympotically_fat]

You’re solving for the function y(x) and by uniqueness they mean does the value of y(0) completely specify y(x)?

[Senegoid]

Generally, when you solve a differential equation, you don’t get a particular number, what you get is a function whose derivative(s) fit the original differential equation you were given.

But wait! Generally, you don’t even get a function – what you get is a whole family of related functions, with a “constant of integration” somewhere in it, commonly denoted by the letter C.

If an initial value is given, you can go a step further, and find one particular one of those functions (that is, one particular value of C). That is what the problem is asking for.

Many differential equations have a “trivial” solution as well as a family of useful solutions. The “trivial” solution is often simply y = 0 (that is, the constant function f(x) = 0 ) or something similar. Typically, these trivial solutions are of no particular interest, and are usually just ignored.

[Budget_Player_Cadet]

@Asympotically_fat

You’re solving for the function y(x) and by uniqueness they mean does the value of y(0) completely specify y(x)?

Ah, so basically “If I have the value of y(0), can I say with complete certainty what c is?” And does c have to be a constant in that case, or can it be dependent on x?

@Senegoid

Generally, when you solve a differential equation, you don’t get a particular number, what you get is a function whose derivative(s) fit the original differential equation you were given.

But wait! Generally, you don’t even get a function – what you get is a whole family of related functions, with a “constant of integration” somewhere in it, commonly denoted by the letter C.

If an initial value is given, you can go a step further, and find one particular one of those functions (that is, one particular value of C). That is what the problem is asking for.

Okay, so far so good. his makes a lot more sense now. I actually wrote down methods for solving for c in more complex cases, and didn’t understand what it meant. Now I get it! Thank you both! But this…

Many differential equations have a “trivial” solution as well as a family of useful solutions. The “trivial” solution is often simply y = 0 (that is, the constant function f(x) = 0 ) or something similar. Typically, these trivial solutions are of no particular interest, and are usually just ignored.

This is something I also don’t get. I understand that y(x) = 0 would be trivial, but how is it a solution if y’ is something like sqrt(abs(x))? Wouldn’t it just be straight-up wrong?

[Asympotically_fat]

If y(x) = 0 for all x, does y’(x) = sqrt(abs(y(x))) for all x?

If so y(x) = 0 is a solution.

[Budget_Player_Cadet]

But… it isn’t. If y(x) is a flat zero, then so is y’(x), and that’s not the same function as sqrt(abs(x)). That’s what’s bothering me.

[friedo]

@ThisUsernameIsForbidden

So hard to read. Is there a board that allows you to embed latex?

The Math Stack Exchange site does, although 99% of the questions there are way beyond my level. I like to peruse it occasionally, though.

[Asympotically_fat]

@Budget_Player_Cadet

But… it isn’t. If y(x) is a flat zero, then so is y’(x), and that’s not the same function as sqrt(abs(x)). That’s what’s bothering me.

The differential equation to be solved isn’t y’(x) = sqrt(abs(x)) though.

[Budget_Player_Cadet]

Well, that’s a derp. Makes sense now. So how about inhomogenous vs. homogenous? Any easy way to tell at a glance which is which? The way proposed to solve inhomogenous involves first finding and solving the homogenous version (for example, going from xy’(x) = y + x^2 to xy’(x) = y and solving that one first to get to the next step); any pointers on how to generalize that? I’m pretty much just getting “the part of the equation with x in it you can’t get on one side” which is pretty weak sauce.

[Thudlow_Boink]

@Asympotically_fat

The differential equation to be solved isn’t y’(x) = sqrt(abs(x)) though.

Right: In case you still don’t see it, it’s y’ = sqrt(abs(y)).

(It might be easier to read if you just wrote y and y’ instead of y(x) and y’(x), and just remembered that x is the independent variable that y is a function of.)

[Thudlow_Boink]

@Budget_Player_Cadet

Well, that’s a derp. Makes sense now.

I’m still a little confused, but if you’re satisfied, that’s good enough.

So how about inhomogenous vs. homogenous? Any easy way to tell at a glance which is which? The way proposed to solve inhomogenous involves first finding and solving the homogenous version (for example, going from xy’(x) = y + x^2 to xy’(x) = y and solving that one first to get to the next step); any pointers on how to generalize that? I’m pretty much just getting “the part of the equation with x in it you can’t get on one side” which is pretty weak sauce.

Here’s how I’d solve your example equation, in case it helps:

xy’ = y + x[sup]2[/sup] is a linear first order differential equation.
It can be put in the form y’ + P(x)y = Q(x) [in this case, y’ - (1/x)y = x]

Multiply through by the integrating factor e^(antiderivative of P(x))
In this case, the antiderivative of –1/x is –(ln x) = ln x[sup]–1[/sup], and e^this = x[sup]–1[/sup].
So, multiplying through by the integrating factor yields x[sup]–1[/sup]y’ – x[sup]–2[/sup]y = 1.

Take advantage of the fact that the left side is the derivative of x[sup]–1[/sup]y to antidifferentiate both sides w.r.t. x, and solve for y. (I get y = x[sup]2[/sup] + cx.)

[Budget_Player_Cadet]

@Thudlow_Boink

I’m still a little confused, but if you’re satisfied, that’s good enough. Here’s how I’d solve your example equation, in case it helps:

xy’ = y + x[sup]2[/sup] is a linear first order differential equation.
It can be put in the form y’ + P(x)y = Q(x) [in this case, y’ - (1/x)y = x]

Multiply through by the integrating factor e^(antiderivative of P(x))
In this case, the antiderivative of –1/x is –(ln x) = ln x[sup]–1[/sup], and e^this = x[sup]–1[/sup].
So, multiplying through by the integrating factor yields x[sup]–1[/sup]y’ – x[sup]–2[/sup]y = 1.

Take advantage of the fact that the left side is the derivative of x[sup]–1[/sup]y to antidifferentiate both sides w.r.t. x, and solve for y. (I get y = x[sup]2[/sup] + cx.)

I understand that method for solving this equation, but how can I generalize that?

[Thudlow_Boink]

@Budget_Player_Cadet

I understand that method for solving this equation, but how can I generalize that?

It depends on what kind of equation you’re trying to generalize it to: first-order or higher-order? linear or non-linear? etc. The method I described should work (theoretucally, assuming you can do the antidifferentiation) on any linear first-order equation.

Here’s a resource you may find helpful: Paul’s Online Math Notes: Differential Equations. Since you asked about distinguishing between homogeneous and nonhomogenous equations, that’s covered on the page on Basic Concepts of Second Order Equations.

The way I’ve learned and taught Diff Eq, I don’t get into the distinction between homogeneous and nonhomogenous equations when dealing with only first-order differential equations, but some books/teachers do. Here’s what seems to be a discussion of first-order differential equations at Khan Academy (haven’t watched it myself).

Also, beware that there is also such a thing as differential equations with homogeneous coefficients, which is something different. I don’t think that’s what you were talking about, but just in case, here’s one description of that.

[MikeS]

@Budget_Player_Cadet

I understand that method for solving this equation, but how can I generalize that?

The method outlined can be used for solve any differential equation of the form y’ + P(x)y = Q(x), regardless of what P(x) and Q(x) are. In fancy math-speak, this equation is first-order (since the highest derivative of y that appears is the first derivative); linear (since all terms that contain y are linear in y); and if Q(x) = 0, then the equation is homogeneous as well.

An equation that is either non-linear or not first-order, however, cannot be solved using the steps that Thudlow Boink has outlined. In such a case, you have to look for other techniques, such as the method of quadrature, separability, or approximation methods such as power-series solutions or the WKB approximation. There’s no one-size-fits-all method for solving differential equations.

[Budget_Player_Cadet]

@Thudlow_Boink

It depends on what kind of equation you’re trying to generalize it to: first-order or higher-order? linear or non-linear? etc. The method I described should work (theoretucally, assuming you can do the antidifferentiation) on any linear first-order equation.

Here’s a resource you may find helpful: Paul’s Online Math Notes: Differential Equations. Since you asked about distinguishing between homogeneous and nonhomogenous equations, that’s covered on the page on Basic Concepts of Second Order Equations.

Wow… That actually makes a lot of sense. That’s the first explanationof differential equations I’ve seen which tries to explain it as though you weren’t already a fucking math whiz. Something our professors could learn something from. :rolleyes: Thanks for the resource.

Also, beware that there is also such a thing as differential equations with homogeneous coefficients, which is something different. I don’t think that’s what you were talking about, but just in case, here’s one description of that.

Never handled that, don’t have the slightest idea what they’re talking about, so if it does appear on the test I’m basically fucked either way. But I guess it’s good to know about.

@MikeS

The method outlined can be used for solve any differential equation of the form y’ + P(x)y = Q(x), regardless of what P(x) and Q(x) are. In fancy math-speak, this equation is first-order (since the highest derivative of y that appears is the first derivative); linear (since all terms that contain y are linear in y); and if Q(x) = 0, then the equation is homogeneous as well.

Well, I’ll write it on the reminder sheet and see if I can’t recognize it. I mean, now I realize that they can be classified in certain ways and that there are certain sure-fire solutions to use, but I can’t for the life of me see which ones we’ve actually covered. I’ll just write down a few of the ones I think will be relevant, because let’s be honest - if the test has second-degree ones, I’m basically screwed either way.

There’s no one-size-fits-all method for solving differential equations.

…Tomorrow’s gonna require a stop by the liquor store, methinks.

Either way, thanks for the help, everyone who’s offered it. I’ll look through your links again, Thudlow, and add a few things to my cheatsheet.

[Budget_Player_Cadet]

Wouldn’t you know it, there wasn’t a single one of these on the makeup exam. Sonovabitch.

Thanks anyways, guys. Passed it with a C-. Not the best grade, but on a class where you’ve got like a 65% failure rate… I’ll take what I can get.

[Senegoid]

@Budget_Player_Cadet

Wouldn’t you know it, there wasn’t a single one of these on the makeup exam. Sonovabitch.

You understand, of course, that your laudable effort in studying these problems for this exam caused that absence of such problems on the exam. (You might have used up the entire supply of available problems.) So, your effort was hardly wasted.

Anyway, just wait until the final.

[Budget_Player_Cadet]

That was the final. Or rather, the retake of the final. And I passed.