Results 1 to 9 of 9

Math Help - Differential Equations Help

  1. #1
    Newbie
    Joined
    Jan 2008
    Posts
    7

    Differential Equations Help

    Hi

    I am having troubles with proving that y=3t+t^2 is one of the solutions of ty'-y=t^2. Can someone show me? I already solved the equation without the book's help but I don't know what to do with it
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by MathIsRearryHard View Post
    Hi

    I am having troubles with proving that y=3t+t^2 is one of the solutions of ty'-y=t^2. Can someone show me? I already solved the equation without the book's help but I don't know what to do with it
    it's easier than it seems

    just find y' and then plug in y and y' into the left hand side of the differential equation and show that you get the right hand side
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by MathIsRearryHard View Post
    Hi

    I am having troubles with proving that y=3t+t^2 is one of the solutions of ty'-y=t^2. Can someone show me? I already solved the equation without the book's help but I don't know what to do with it
    I'm confused about something. You say you solved the differential equation, but you don't know if the given solution is one of them? I would think you'd simply be able to compare them.

    Anyway, proving the given solution solves the differential equation is easy: Simply stick the solution into the equation:
    ty^{\prime} - y = t^2
    and see if it is an identity.

    y = 3t + t^2

    y^{\prime} = 3 + 2t

    So
    t(3 + 2t) - (3t + t^2)

    = 3t + 2t^2 - 3t - t^2

    = t^2
    as desired.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2008
    Posts
    7
    Thanks guys. I was overthinking it too much.

    I solved the equation and I got this wierd answer (?):

    y= -t-1 +Ce^t

    How is it possible to substitute with this if we don't know what "C" is? I just wanna know
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by MathIsRearryHard View Post
    Thanks guys. I was overthinking it too much.

    I solved the equation and I got this wierd answer (?):

    y= -t-1 +Ce^t

    How is it possible to substitute with this if we don't know what "C" is? I just wanna know
    how exactly did you get that solution. i did it and got y = t^2 + Ct, which is the correct form.

    (we use the integrating factor method here)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2008
    Posts
    7
    Quote Originally Posted by Jhevon View Post
    how exactly did you get that solution. i did it and got y = t^2 + Ct, which is the correct form.

    (we use the integrating factor method here)
    ty'-y+t^2

    Ok, this is my first time dealing with equations like these so I might be wrong. Here's how I did it. (A little extra info about my background: I've only been exposed to these type of problems for 2 1/2 days)

    The integrating factor here is e^-t since p(t)= -1 and integrating you would get -t.

    So,

    (ye^-t)'= te^-t

    u=t du= dx dv=e^-t dx

    Integrating both sides, I finally get

    y= -t- 1 +Ce^t

    Please tell me if this is wrong when you see it
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by MathIsRearryHard View Post
    ty'-y+t^2

    Ok, this is my first time dealing with equations like these so I might be wrong. Here's how I did it. (A little extra info about my background: I've only been exposed to these type of problems for 2 1/2 days)

    The integrating factor here is e^-t since p(t)= -1 and integrating you would get -t.

    So,

    (ye^-t)'= te^-t

    u=t du= dx dv=e^-t dx

    Integrating both sides, I finally get

    y= -t- 1 +Ce^t

    Please tell me if this is wrong when you see it
    to apply the integrating factor method, it's best, if not required, to have the coefficient of y' to be 1. so we would divide through by t before doing anything.

    you should have noticed what you had does not work. since if we multiply through by e^{-t} we get:

    te^{-t}y' - e^{-t}y = t^2e^{-t}

    clearly the left hand side is not the derivative of something using the product rule. the te^{-t} would ensure that the derivative te^{-t}y has three terms when fully expanded.

    however, if we divide by t first, we have y' - \frac 1ty = t

    then \mu (t) = exp \left( - \int \frac 1t~dt \right) = \frac 1t, multiplying through we get:

    \frac 1ty' - \frac 1{t^2}y = 1

    and now the left side is the derivative given by the product rule of \frac 1ty

    and we're in business
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2008
    Posts
    7
    I understand now how you got that but I'm a little shaky with these IF's because I tend to look at them by using the formula e^integral p(t) dx. I did divide by t on both sides before I multiplied the IF on both sides.

    Thanks for the help. That is certainly something I've never seen before.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by MathIsRearryHard View Post
    I understand now how you got that but I'm a little shaky with these IF's because I tend to look at them by using the formula e^integral p(t) dx. I did divide by t on both sides before I multiplied the IF on both sides.
    i'm not sure what's your hangup. just put it in your mind that the ideal form to have is y' + p(t)y = g(t), here exp \left(  \int p(t)~dt \right) is the integrating factor, and you are to multiply through the whole equation by it.

    see post #21 here
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 'Differential' in differential equations
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: October 5th 2010, 11:20 AM
  2. Replies: 2
    Last Post: May 18th 2009, 04:49 AM
  3. Differential Equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 12th 2009, 06:44 PM
  4. Replies: 5
    Last Post: July 16th 2007, 05:55 AM
  5. Replies: 3
    Last Post: July 9th 2007, 06:30 PM

Search Tags


/mathhelpforum @mathhelpforum