newbie DE question: "y'' + y = o is important because it appears in physics." How?

Vol

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newbie DE question: "y'' + y = o is important because it appears in physics." How?

A differential equation like y'' + y = o is important because it appears in physics. I am confused. How does it appear in physics? Because doesn't y stand for f(x) which can be any function? like x^2 + 1 So, without knowing the actual function in terms of x, how does solving y'' + y = 0 help? I know this is probably a bad question. But I am a newbie. Do you plug in f(x) after you solve for y?
 
A differential equation like y'' + y = o is important because it appears in physics. I am confused. How does it appear in physics? Because doesn't y stand for f(x) which can be any function? like x^2 + 1 So, without knowing the actual function in terms of x, how does solving y'' + y = 0 help? I know this is probably a bad question. But I am a newbie. Do you plug in f(x) after you solve for y?

1) "plug in" doesn't mean anything.
2) This is the beauty of Differential Equations. You do not need to know the nature of the function to have a conversation about it.
3) Often time, yes, you do find an actual function and substitute it for y. This is known as a demonstration or usually, in a problem statement, it is said, "demonstrate that" or "show that".
 
To understand the workings behind solving basic ODE's some linear algebra knowledge is necessary. A homogeneous equation like the one you posted can only have two linearly independent general solutions that form to give the general solution. Once you find two of them, any two, you have solved the equation.

To solve

\(\displaystyle y'' + y=0\)

means to find (all) functions \(\displaystyle y=f(x)\) (or \(\displaystyle f(t), etc\)) whose acceleration is equal to the negative of the functions value at any give input/time.

You may remember two such functions from calculus: \(\displaystyle f_1(x) = \sin x, f_2(x)=\cos x\). These turn out to be linearly independent, and so the general solution is the family of functions given by \(\displaystyle f(x)=c_1\sin x + c_2\cos x\) where \(\displaystyle c_1,c_2\) are arbitrary numbers. This is equivalent to the solution of \(\displaystyle g(x)=2c_1\cos(x)-\pi c_2(\sin(x) - 4\cos(x))\) which is needlessly complicated but technically valid.

A typical course will lend you a healthy serving of techniques to solve these and more complicated types of differential equations
 
Vol

1) "plug in" doesn't mean anything.
2) This is the beauty of Differential Equations. You do not need to know the nature of the function to have a conversation about it.
3) Often time, yes, you do find an actual function and substitute it for y. This is known as a demonstration or usually, in a problem statement, it is said, "demonstrate that" or "show that".

How is it you do not need to know the nature of the function? Could you demonstrate? Or anybody?
 
How is it you do not need to know the nature of the function? Could you demonstrate? Or anybody?

Well, the differential equation itself tells you something about the nature of the function.

Simply stating the equation, y" + y = 0 might mean simply, "There is a function..." or "We hope there is a function..." At this point, what is it that you know? Virtually nothing. And yet, there it is.

If we then add insight, or construe the equation to mandate that we find a solution, then that's another matter.

One might say

y" + y = 0

Okay, so there is a function that when added to its 2nd derivative gives zero.

Another might say, "Oh, is there?".

And the investigation is afoot.

Your function: y = x^2 + 1, y' = 2x, y" = 2 ==> 2 + (x^2 + 1) = x^2 + 3 and that is NOT zero ANYWHERE in the Real Plane. Thus the form of y that IS a solution remains a mystery.
 
A differential equation like y'' + y = o is important because it appears in physics. I am confused.

How does it appear in physics?

Because doesn't y stand for f(x) which can be any function? like x^2 + 1

So, without knowing the actual function in terms of x, how does solving y'' + y = 0 help?

Do you plug in f(x) after you solve for y?

1. An equation like this might be used in physics with y representing (hypothetically) a distance traveled, or an electrical charge, or just about anything. Much of physics involves relationships between quantities and their derivatives. (If you are asking about this specific equation, we could search for it.)
2. The variable/function y stands for a particular function that satisfies the equation, just as in algebra x stands for a particular number that satisfies an equation. It is not any function at all.
3. Solving the differential equation means finding the actual function in terms of x. The goal is to know the function!
4. After you have solved, you will have the function f(x), and can check that it does satisfy the equation by "plugging it in", as you do in algebra.

Does this answer your questions?
 
1. An equation like this might be used in physics with y representing (hypothetically) a distance traveled, or an electrical charge, or just about anything. Much of physics involves relationships between quantities and their derivatives. (If you are asking about this specific equation, we could search for it.)
2. The variable/function y stands for a particular function that satisfies the equation, just as in algebra x stands for a particular number that satisfies an equation. It is not any function at all.
3. Solving the differential equation means finding the actual function in terms of x. The goal is to know the function!
4. After you have solved, you will have the function f(x), and can check that it does satisfy the equation by "plugging it in", as you do in algebra.

Does this answer your questions?


I'll have to study further and think about it. Thank you all for your reply. I will be back or post a new thread with more questions.
 
Vol

OK. I get it. You have to solve for y by finding f(x), which is sin or cos in this case. But how did you find sin or cos? And what if we were to find it using a series solution?
 
OK. I get it. You have to solve for y by finding f(x), which is sin or cos in this case. But how did you find sin or cos? And what if we were to find it using a series solution?

Well, presumably that is what your course is about!

The answer to the question, "How can you find a solution?" may take a whole textbook, depending on the particular kinds of differential equations you are asking about. In some cases, the book will essentially just say, if the equation looks like this, then the answer looks like that (which may have once been discovered almost by accident!). To some extent, a course on differential equations amounts to a collection of special tricks that have been found useful for various types (and there are whole courses on different types, such as "ordinary" vs. "partial").
 
A differential equation like y'' + y = o is important because it appears in physics. I am confused. How does it appear in physics?
I presume, since you are asking about differential equations, that you have taken calculus and know that y'' is the second derivative. If, for example, y is the position of some object, say its distance from a given point, then y', the first derivative, is the rate of change of position- the speed at which that object is moving. And y'', the second derivative, is the rate at which that speed is changing- the acceleration of the object. Second order differential equations are especially important in physics because "Force equals mass times acceleration". Taking mass to be 1, y''+ y= 0, y''= -y, means that the acceleration, and so the force is the negative of the distance- a spring is an example of such a force- the more it is stretched the more the force is back the other way.

Because doesn't y stand for f(x) which can be any function? like x^2 + 1
Well y can be any function that satisfies that equation! No y cannot be \(\displaystyle x^2+ 1\) because if \(\displaystyle y= x^2+ 1\), then \(\displaystyle y'= 2x\) and \(\displaystyle y''= 2\). \(\displaystyle y''+ y= 2+ x^2+ 1\) is not equal to 0.

So, without knowing the actual function in terms of x, how does solving y'' + y = 0 help? I know this is probably a bad question. But I am a newbie. Do you plug in f(x) after you solve for y?
I am tempted to say that you start by taking a Calculus class!

One of the things that you learn in a Calculus class is to take derivatives and, in particular, that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x). So if y= sin(x) then y''= -sin(x)= -y. Similarly if y= cos(x) then y''= -cos(x)= -y. We also know, from the general theory of linear equations, that any linear combination of solutions is also a solution: The general solution to this differential equation is y(x)= Acos(x)+ Bsin(x).
 
I feel compelled to respond to "A differential equation like y'' + y = o is important because it appears in physics." Well, if you are a physicist or especially like physics that might be a reason this equation is important. But the crucial thing about mathematics is that it can be applied to many different fields. If, for example, you were a sociologist interested in "population" trends then y might be the population in a specific area. The simplest such equation would be y'= ky so that the rate at which population increases is proportional to the current population. The general solution to such an equation is \(\displaystyle y(t)= Ce^{kt}\) so that, for positive k, the population would increase, very rapidly without bound. But there might well be other restrictions, such as food limits that change that equation. One could imagine a situation where it is not the rate of increase of population but the rate of change of that rate of increase that is proportional to the negative of the population. In that case we would have the equation y''= -y or y''+ y= 0. As I said before, the general solution is y(t)= Acos(t)+ Bsin(t) so that population increases, then decreases, then increases again.
 
Math course

But I heard that if you take a math course they do not have hours during working hours. I also heard that if you flunk the course they will not give you a refund. :confused:
 
But I heard that if you take a math course they do not have hours during working hours. I also heard that if you flunk the course they will not give you a refund. :confused:
Who is "they"? And how do these questions relate to the original post? Please be complete. Thank you! ;)
 
Vol

I am confused as to when you can and cannot use a series solution to solve a differential equation. Can you apply a series solution to say solve a differential mixing problem?:confused:
 
The crucial thing is that the equation be linear so that you can add different solutions to get a new solution.
 
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