Why Is the Interval of Solution (0, inf) Instead of (-inf, inf)?

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In summary: No, you don't have to take into account every equation. This is just a differential equation.In summary, x\frac {dy}{dx} - y = x^2sin(x) has a domain of (-\infty,\infty) and a solution of (0,inf). There are no transient terms in the solution.
  • #1
Rijad Hadzic
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Homework Statement


Find the general solution of the given differential equation. Give the largest interval I over which the interval is defined.Determine wether there are any transient terms in the general solution11.

[itex] x\frac {dy}{dx} - y = x^2sin(x)[/itex]

Homework Equations

The Attempt at a Solution


[itex] \frac {dy}{dx} + x^{-1}y = sin(x) [/itex]

[itex] e^{-\int {x^{-1} dx }} = 1/x [/itex]

[itex] d/dx[yx^{-1}] = sin(x) [/itex]

[itex] yx^{-1} = -cos(x) + c [/itex]

[itex] y = -xcos(x) + cx [/itex]

I don't understand why the interval of solution isn't (-inf, inf), as that would be the largest solution. My book says its (0,inf)

How could zero not be included?

Why would the interval of solution be (0,inf)? Couldn't (-inf,inf) be a solution as well? Because even if x = 0, you simply get -(0)(1) + c(0) = 0, which is a real number, so why can't it be (-inf,inf)??
 
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  • #2
Rijad Hadzic said:

Homework Statement


Find the general solution of the given differential equation. Give the largest interval I over which the interval is defined.Determine wether there are any transient terms in the general solution11.

[itex] x\frac {dy}{dx} - y = x^2sin(x)[/itex]

Homework Equations

The Attempt at a Solution


[itex] \frac {dy}{dx} + x^{-1}y = sin(x) [/itex]

[itex] e^{-\int {x^{-1} dx }} = 1/x [/itex]
Your integrating factor is wrong. Never mind. After noticing your other error Mark44 has already mentioned it.
 
  • #3
Rijad Hadzic said:
[itex] x\frac {dy}{dx} - y = x^2sin(x)[/itex]


[itex] \frac {dy}{dx} + x^{-1}y = sin(x) [/itex]
Besides what Dick said, you have two mistakes going from the first equation above to the second.
 
  • #4
Sorry guys. I cannot for the life of me find an edit button on the OP so I'll just post here.

I copied my work down wrong. I was tired af when I did all the itex crap but I had to solution right on my paper, just not the interval of solution part (I got different answer than the book on that part)

so you have

[itex] x\frac {dy}{dx} -y = x^2sin(x) [/itex]

[itex] \frac {dy}{dx} - \frac 1xy = xsin(x) [/itex]

[itex] e^{-\int {\frac 1x dx}} [/itex] = [itex] 1/x [/itex]

[itex] \frac {d}{dx} [\frac 1x y] = -\frac 1{x^2}y + \frac 1x \frac {dy}{dx} [/itex] = [itex] sin(x) [/itex]

take integral from both sides

[itex] -xcos(x) + cx = y [/itex]

this is my solution.

I don't understand why the integral isn't (-inf,inf), instead its (0,inf)

1)Why would zero not be included.
2)Why wouldn't we take the largest possible interval ( (-inf,inf)) ??
 
  • #5
Rijad Hadzic said:
Sorry guys. I cannot for the life of me find an edit button on the OP so I'll just post here.

I copied my work down wrong. I was tired af when I did all the itex crap but I had to solution right on my paper, just not the interval of solution part (I got different answer than the book on that part)

so you have

[itex] x\frac {dy}{dx} -y = x^2sin(x) [/itex]

[itex] \frac {dy}{dx} - \frac 1xy = xsin(x) [/itex]

[itex] e^{-\int {\frac 1x dx}} [/itex] = [itex] 1/x [/itex]

[itex] \frac {d}{dx} [\frac 1x y] = -\frac 1{x^2}y + \frac 1x \frac {dy}{dx} [/itex] = [itex] sin(x) [/itex]

take integral from both sides

[itex] -xcos(x) + cx = y [/itex]

this is my solution.

I don't understand why the integral isn't (-inf,inf), instead its (0,inf)

1)Why would zero not be included.
2)Why wouldn't we take the largest possible interval ( (-inf,inf)) ??
1) Because your DE in standard form (the form that is used in theorems about existence and uniqueness) is dy/dx + f(x)y = g(x). Here, f(x) = -1/x, which is defined on either ##(-\infty, 0)## or on ##(0, \infty)##. Zero is not included in either interval.
2) Because that interval includes 0, which is not in the domain of f, as I described it above.

For this problem, with no initial condition, I think you could pick either of the half-lines that I wrote. If an initial condition is given, though, then you have to choose the interval that includes your initial condtion. For example, if y(1) = 2, you would choose the positive interval.
 
  • #6
I don't agree with Mark44 about this. Assuming the DE was presented in the form the OP gave:$$
x\frac {dy}{dx} -y = x^2\sin(x)$$ I don't see any reason to use any theory or theorems. Direct substitution shows the solution solves the DE on ##(-\infty,\infty)##. If the DE was originally given in the form$$
\frac {dy}{dx} -\frac 1 x y = x\sin(x)$$then ##x=0## can't be in the domain of any solution. The two forms of the DE aren't exactly equivalent.
 
  • #7
Mark, your answer does make sense to me. I was only looking at the solution, and not when it was a differential equation, but looking back 1/x is undefined at x=0 which makes sense to me now.

LCKurtz, I thought you have to take into account every equation that you would have? For example, you have f(x), then f'(x) etc and the interval of solution has to be valid for both of them.
 
  • #8
Rijad Hadzic said:
LCKurtz, I thought you have to take into account every equation that you would have? For example, you have f(x), then f'(x) etc and the interval of solution has to be valid for both of them.
I'm not sure I understand what you are saying here. Maybe this example will help. Say you want the general solution to ##y' = y##. You might try to solve it by separation of variables like this:$$
\frac {dy} y = 1dx$$ $$\ln |y| = x+ C$$ $$|y| = e^{x + C} = e^x e^C = Ke^x$$ with ##K>0##. You could rewrite this as $$
y = \pm Ke^x = De^x$$ where ##D \ne 0##. Notice that this solution does not include ##y= 0##, which is no surprise since we divided by ##y## in the first step of separation of variables. Yet ##y=0## is a solution of the original DE ##y'=y##. The reason the solution ##y=0## was missed was because of the method used to solve the DE. But if you solve the original DE with an integrating factor you will get the general solution which is ##y = Ce^x## for any ##C##. I'm not sure whether this helps you or not, just my 2 cents worth on the subject.
 
  • #9
Actually it does help me and I am following what you are saying, and at the same time, it is making me even more confused, lol.

That's why I think Mark brought up the form used in theorems about existence and uniqueness. How exactly are we suppose to be able to know which interval to use? I think that's why he brought up "theorems about existence and uniqueness."

Again this is my interpretation, I would like to see what Mark says about this. As it stands I'm divided between both of your answers and I would like to get to the bottom of this..

If it helps any my books answer was (0,inf) but again the book could be wrong (wouldn't be the first time)
 
  • #10
Here's a theorem from one of my DE textbooks, "A First Course in Differential Equation," by Frank G. Hagin.
Let p and q be continuous for ##\alpha < x < \beta##. Then the initial value problem
##y' = py + q, y(x_0) = y_0## has a solution (and only one), expressed by
##y(x) = \frac 1 {\phi(x)}\left[ y(x_0) + \int_{x_0}^x p(t)dt\right]##
where ##\phi(x) = \exp\left[-\int_{x_0}^x p(t)dt\right]##
Here p and q are functions of x, but your problem is not an initial value problem, so this theorem isn't an exact fit for your problem.
Your DE, written in this form, is ##y' = \frac 1 x y + x\sin(x)##, so p(x) of the theorem is ##p(x) = \frac 1 x##, and this function is defined either on ##(-\infty, 0)## or on ##(0, \infty)##. At this point, without an initial condition being given, you could choose either of these intervals. If an initial condition had been given, say, ##y(x_0) = y_0##, the sign of ##y_0## would determine which of the two intervals was the appropriate one. Note that you could not have ##y(x_0) = 0##.

I should also mention that, to get your integrating factor, you had to divide through by x, meaning that whatever interval you have for your solution can't x = 0.
 
  • #11
But my point was that the form in which the equation is presented or the method of solution may affect the answer. And the OP has not verified for us whether the equation was actually given in the form ##xy' - y = x^2\sin x## as he posted. The thing is, although the existence and uniqueness theorem doesn't guarantee a solution on ##(-\infty,\infty)## for the equation in this form, the fact is that it does have such a solution given by ##y=cx - x\cos x##.
 

Related to Why Is the Interval of Solution (0, inf) Instead of (-inf, inf)?

What is an interval of solution?

An interval of solution is a range of values that satisfy a given equation or inequality. It represents all possible solutions to the equation within that range.

How do you find the interval of solution for an equation?

To find the interval of solution for an equation, you can graph the equation and determine where it intersects the x-axis. The x-values of these intersections represent the interval of solution. Alternatively, you can solve the equation algebraically and use the solution as the interval of solution.

What is the difference between an open and closed interval of solution?

An open interval of solution does not include the endpoints, while a closed interval of solution includes the endpoints. For example, an open interval of solution would be (0,5), while a closed interval of solution would be [0,5].

Can an equation have multiple intervals of solution?

Yes, an equation can have multiple intervals of solution. This occurs when there are multiple x-intercepts on the graph of the equation or when there are multiple solutions to the equation.

How can you check if a value is within the interval of solution for an equation?

You can check if a value is within the interval of solution by plugging it into the equation and seeing if it satisfies the given equation. If the value makes the equation true, then it is within the interval of solution.

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