Results 1 to 8 of 8

Math Help - Problem with a differential equation

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Problem with a differential equation

    I'm having some trouble solving a simple differential equation, and yes I'm a bit lazy to check my notes.
    It's a simple one, here it is : We have that x'=f(t,x), x(0)=0. Use the fact that dt/dx=(dx/dt)^{-1} when dx/dt≠0.
    f(t,x)=x^{-2}.
    It's certainly a beginner question but I have a doubt in this case : x'=dx/dt? I guess yes, since otherwise I don't see what could be x'!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by arbolis View Post
    I'm having some trouble solving a simple differential equation, and yes I'm a bit lazy to check my notes.
    It's a simple one, here it is : We have that x'=f(t,x), x(0)=0. Use the fact that dt/dx=(dx/dt)^{-1} when dx/dt≠0.
    f(t,x)=x^{-2}.
    It's certainly a beginner question but I have a doubt in this case : x'=dx/dt? I guess yes, since otherwise I don't see what could be x'!
    Yes, x' is an alternative notation for dx/dt.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Still in the mud

    Thanks Opalg. This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
    Also... well I need to read much more on calculus, but why can we separate the dt from the expression dx/dt? It's not a division...
    I think I can solve problems like this one : x'=t^3 and with the same initial condition as I posted at first. But when I have x' in function of x, then I'm lost. (that's why I'm posting here right now).
    It's also confusing to me that instead of writing x'(t)=t for example, we just write x'=t.
    What I've thought about the problem is \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow dx=\frac{1}{x^2} dt. Now from there should I integrate??? I will have x= something depending of x but with dt at the end. So I guess this is not the way to do it.
    Can someone give me a hint solving the differential equation?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by arbolis View Post
    Thanks Opalg. This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
    Also... well I need to read much more on calculus, but why can we separate the dt from the expression dx/dt? It's not a division...
    I think I can solve problems like this one : x'=t^3 and with the same initial condition as I posted at first. But when I have x' in function of x, then I'm lost. (that's why I'm posting here right now).
    It's also confusing to me that instead of writing x'(t)=t for example, we just write x'=t.
    What I've thought about the problem is \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow dx=\frac{1}{x^2} dt. Now from there should I integrate??? I will have x= something depending of x but with dt at the end. So I guess this is not the way to do it.
    Can someone give me a hint solving the differential equation?
    You're almost there. This differential equation is separable. Just take \frac{dx}{dt} = \frac{1}{x^2} and multiply both sides of the equation by x^2 dt to yield x^2 dx = dt. Now you can integrate both sides.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by arbolis View Post
    This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
    That's the trouble with the x' notation: it doesn't make explicit what the variable is. In this case, it is implicitly understood that x is a function of the variable t.

    In the same way, if you see the notation y', it will usually mean dy/dx, but you have to look at the rest of the problem to see whether the independent variable is in fact x (or it may be t, or some other letter).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    I think I made an error. Here is what I've done :
    \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow \int x^2 dx=\int dt\Leftrightarrow x=(3(t-C))^{1/3} soon after this we realize that C which is the constant produced when integrating is equal to 0. So at the end I found that x(t)=(3t)^{1/3}. Is it right?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by arbolis View Post
    I think I made an error. Here is what I've done :
    \frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow \int x^2 dx=\int dt\Leftrightarrow x=(3(t-C))^{1/3} soon after this we realize that C which is the constant produced when integrating is equal to 0. So at the end I found that x(t)=(3t)^{1/3}. Is it right?
    If you can assume that C is equal to zero, then that answer is correct.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Thanks both for your valuable help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 11th 2011, 01:17 AM
  2. Another differential equation problem
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: January 5th 2011, 12:50 AM
  3. Differential Equation Problem 2
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: September 28th 2009, 12:31 AM
  4. differential equation problem
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: February 20th 2009, 02:01 AM
  5. Differential Equation Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 23rd 2007, 05:30 AM

Search Tags


/mathhelpforum @mathhelpforum