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Math Help - Second order ODE

  1. #1
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    Second order ODE

    Hi, I am trying to solve the following differential equation, i'm not sure what my best method is.

    2*sinh(x)*y'' - 2*cosh(x)*y' + y*sinh(x)^3 = 0, y(1) = 0, y(2) = 2,

    I know what the answer is as I could put it into Maple and it's produced, but im not sure of the intermediate steps.

    Thanks
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  2. #2
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    Re: Second order ODE

    You can create a system of linear first-order equations by setting z(x) = y'(x). But it's probably going to be better to just guess at the solution - something like y(x) = A cosh^n(x) + B sinh^m(x) might work.

    We have a differential equations forum here - you might have better luck posting your questions there.

    - Hollywood
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  3. #3
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    Re: Second order ODE

    Thanks. but i'm really not sure how to set this problem out to solve. Usually on my course the material just involves finding the Euler-Lagrange equation, but we never actually have to solve the differential equations.....

    Any additional help would be much appreciated to get going on this

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  4. #4
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    Re: Second order ODE

    Making the change of variable

    t=coshx

    gets you the equation

    2\frac{d^{\,2}y}{dt^{\,2}}+y=0

    which you solve, back substitute, and then evaluate the constants.
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  5. #5
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    Re: Second order ODE

    Thanks, I can solve that easy enough to get t(x) =

    c\cos \left( 1/2\,x\sqrt {2} \right) +k\sin \left( 1/2\,x\sqrt {2}\right)

    I just dont understand what exactly im meant to be doing with this solution, what exactly im meant to substitute it into and as a subsitutefor what etc....?

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  6. #6
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    Re: Second order ODE

    To put it in the same notation as BobP used, y(t)=c_1\cos{\frac{t}{\sqrt{2}}}+c_2\sin{\frac{t}{  \sqrt{2}}}. Now you need to substitute t=\cosh{x} and evaluate the constants c_1 and c_2.

    - Hollywood
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  7. #7
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    Re: Second order ODE

    Right, I see, so I would then have something like this. I'm sorry, this is probably very easy, just i'v never dealt with this before so just clarifying the method.

    y \left( coshx \right) =c\cos \left( 1/2\,\sqrt {2}\cosh \left( x \right) \right) +k\sin \left( 1/2\,\sqrt {2}\cosh \left( x \right)  \right)

    Do I need to use the boundary conditions at all?

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  8. #8
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    Re: Second order ODE

    yes.
    you have two unknown constants, c and k, in your solution.
    you are given two boundary conditions.
    Use the boundary condition to determine values for c and k and you will be finished.

    Also, your solution is

    $$y(x)=c \cos\left(\frac{\cosh(x)}{\sqrt{2}}\right) + k \sin\left(\frac{\cosh(x)}{\sqrt{2}}\right)$$

    not

    $$y(\cosh(x))= \dots$$

    finally, just as a matter of style, constants are usually called c1, c2, etc. Using c and k isn't incorrect, just odd and potentially confusing.
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  9. #9
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    Re: Second order ODE

    Thanks romsek, that definitely helps. I'm told however to show that it equals the following: (see image)

    Second order ODE-odemhf.png

    I can see how I can evaluate the constants c1 and c2 like you said, but that doesnt help me (as far as my tiny brain can see) get it to the above format....?

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  10. #10
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    Re: Second order ODE

    Quote Originally Posted by sirellwood View Post
    Thanks romsek, that definitely helps. I'm told however to show that it equals the following: (see image)

    Click image for larger version. 

Name:	ODEMHF.png 
Views:	3 
Size:	7.0 KB 
ID:	30253

    I can see how I can evaluate the constants c1 and c2 like you said, but that doesnt help me (as far as my tiny brain can see) get it to the above format....?

    Thanks
    well... There's no magic that I know of. I was able to whip through the algebra and trig in Mathematica w/o too much trouble but by hand it would be a nightmare.

    I attached an image of the Mathematica screen but realy if you want to get good at this sort of stuff you have to push your pencil through it yourself.

    The identity sin(a-b)=sin(a)cos(b)-cos(b)sin(a) plays a big part in simplifying it.

    Second order ODE-clipboard01.jpg
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