Results 1 to 12 of 12

Math Help - [SOLVED] How to answer this question

  1. #1
    Member
    Joined
    May 2009
    Posts
    99

    [SOLVED] How to answer this question

    Show that y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 - 2y\frac{dy}{dx} = 4

    given that y = (e^{2x} - 4x + 3)^{\frac{1}{2}}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,507
    Thanks
    1403
    Quote Originally Posted by mark1950 View Post
    Show that y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 - 2y\frac{dy}{dx} = 4

    given that y = (e^{2x} - 4x + 3)^{\frac{1}{2}}
    Can you work out \frac{dy}{dx}?

    Or \frac{d^2y}{dx^2}?


    Can you substitute these into the DE?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    99
    Yes, I can but what I got was a very, very long equation.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,507
    Thanks
    1403
    Quote Originally Posted by mark1950 View Post
    Yes, I can but what I got was a very, very long equation.
    Well that's how it's solved, so do it.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mark1950 View Post
    Yes, I can but what I got was a very, very long equation.
    Please post your work so that it can be reviewed. Be sure to include your answer for \frac{dy}{dx} and \frac{d^2y}{dx^2}.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2009
    Posts
    99
    @Prove it

    Dude, if I knew how to solve it, I would not have posted this question here.

    @Mr. Fantastic

    Thanks. After this step, I simply do not know how to continue. I tried substituting dy/dx into them but simply couldn't get rid of the exponential, e so as to prove that its right. You know what I mean, right?

    \frac{dy}{dx} <br />
= \frac{1}{2}(e^{2x} - 4x + 3)^{-\frac{1}{2}}(2e^{2x} - 4)
    = \frac{e^{2x} - 2}{\sqrt{e^{2x} - 4x + 3}}
    \frac{d^2y}{dx^2}<br />
= \frac{(e^{2x} - 4x + 3)(2e^{2x}) - (e^{2x} - 2)^2}{\sqrt{(e^{2x} - 4x + 3)^3}}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,507
    Thanks
    1403
    Quote Originally Posted by mark1950 View Post
    Show that y\frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 - 2y\frac{dy}{dx} = 4

    given that y = (e^{2x} - 4x + 3)^{\frac{1}{2}}
    So now plug everything you've found into

    y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 - 2y\frac{dy}{dx}

    and simplify.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mark1950 View Post
    @Prove it

    Dude, if I knew how to solve it, I would not have posted this question here.

    @Mr. Fantastic

    Thanks. After this step, I simply do not know how to continue. I tried substituting dy/dx into them but simply couldn't get rid of the exponential, e so as to prove that its right. You know what I mean, right?

    \frac{dy}{dx} <br />
= \frac{1}{2}(e^{2x} - 4x + 3)^{-\frac{1}{2}}(2e^{2x} - 4)
    = \frac{e^{2x} - 2}{\sqrt{e^{2x} - 4x + 3}}
    \frac{d^2y}{dx^2}<br />
= \frac{(e^{2x} - 4x + 3)(2e^{2x}) - (e^{2x} - 2)^2}{\sqrt{(e^{2x} - 4x + 3)^3}}
    Your derivatives are OK. Now the unfortunate fact is that you need to substitute y and the derivatives into the left hand side of the differential equation. Then simplify it (hopefully to 4). This will take a fair bit of algebra but is not difficult - just very tedious (hence my use above of the word unfortunate).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,507
    Thanks
    1403
    Quote Originally Posted by mr fantastic View Post
    Your derivatives are OK. Now the unfortunate fact is that you need to substitute y and the derivatives into the left hand side of the differential equation. Then simplify it (hopefully to 4). This will take a fair bit of algebra but is not difficult - just very tedious (hence my use above of the word unfortunate).
    My CAS tells me the answer is, indeed, 4.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Prove It View Post
    My CAS tells me the answer is, indeed, 4.
    It's always good to know that what you're trying to do can be done (there's always a niggling suspicion that the given solution is wrong in problems like this ....)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,507
    Thanks
    1403
    Quote Originally Posted by mr fantastic View Post
    It's always good to know that what you're trying to do can be done (there's always a niggling suspicion that the given solution is wrong in problems like this ....)
    Indubidably.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    May 2009
    Posts
    99
    Oh darn...I was looking for a simpler way but nvm...there are some things that can't be simplified. Thanks, anyway for the assurance that my derivatives are correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: May 25th 2009, 09:07 AM
  2. [SOLVED] which is the correct answer ?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 5th 2009, 08:45 AM
  3. Replies: 1
    Last Post: August 29th 2008, 10:17 AM
  4. [SOLVED] Have I got the right answer?
    Posted in the Algebra Forum
    Replies: 0
    Last Post: July 26th 2008, 06:37 AM
  5. Replies: 3
    Last Post: April 1st 2008, 10:07 PM

Search Tags


/mathhelpforum @mathhelpforum