Results 1 to 10 of 10

Thread: Linear Second Order DE

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    10

    Linear Second Order DE

    The initial question was :

    y''-10y'+61y=0

    i've got the general solution which is e^5x(A cos(6x)+b sin(6x))

    given that the initial conditions were y(0)=0 and y'(0)=-9,find A and B

    after i calculated shudn't A =0?it seems i was wrong?

    Please help thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Ted
    Ted is offline
    Member
    Joined
    Feb 2010
    From
    China
    Posts
    240
    Thanks
    12
    A=0 is correct.

    What makes you think it is wrong?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    $\displaystyle \displaystyle A = 0$ is correct.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2011
    Posts
    10
    then how would the general solution be like?as in y(x)=?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Differentiate it and substitute your other condition to solve for $\displaystyle \displaystyle B$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2011
    Posts
    10
    I've got B= 54

    thus my equation should be 'exp(5*x)*(54*sin(6*x))'..but somehow its wrong
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    I don't know how you can get $\displaystyle \displaystyle B = 54$...

    $\displaystyle \displaystyle y'(x) = 5Be^{5x}\sin{(6x)} + 6Be^{5x}\cos{(6x)}$, and since $\displaystyle \displaystyle y'(0) = -9$

    $\displaystyle \displaystyle -9 = 5Be^{5\cdot 0}\sin{(6\cdot 0)} + 6Be^{5\cdot 0}\cos{(6\cdot 0)}$

    $\displaystyle \displaystyle -9 = 6B$

    $\displaystyle \displaystyle B = -\frac{9}{6} = -\frac{3}{2}$...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2011
    Posts
    10
    Ahh..i integrated it wrongly!!Thanks alot!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by topsquark View Post
    You made a slight error. A = 0, so the sine term should not be there. It worked because sin(6*0) = 0.

    -Dan
    You need to use the product rule to differentiate $\displaystyle \displaystyle y(x) = Be^{5x}\sin{(6x)}$.

    $\displaystyle \displaystyle u(x) = Be^{5x} \implies u'(x) = 5Be^{5x}$ and $\displaystyle \displaystyle v(x) = \sin{(6x)} \implies v'(x) = 6\cos{(6x)}$.

    Therefore $\displaystyle \displaystyle y'(x) = u'(x)v(x) + u(x)v'(x)$

    $\displaystyle \displaystyle = 5Be^{5x}\sin{(6x)} + 6Be^{5x}\cos{(6x)}$.


    I did not make a mistake.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by Prove It View Post
    I did not make a mistake.
    I....ummmm...I was just testing you. Yeah, that's it. Testing. (crawls away into a dark corner.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear program with higher order non-linear constraints.
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: Sep 12th 2010, 02:36 AM
  2. Help with 2nd order non-linear DE
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Mar 24th 2010, 01:12 PM
  3. Replies: 4
    Last Post: Aug 12th 2008, 04:46 AM
  4. Help with third order, non-linear DE.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Aug 9th 2008, 03:45 PM
  5. Replies: 1
    Last Post: May 11th 2007, 03:01 AM

Search Tags


/mathhelpforum @mathhelpforum