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Math Help - Variation of Parameters

  1. #1
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    Variation of Parameters

    I have a problem understanding a certain step using the Variation of Parameters Method.

    take for example

    y''+y=sec(x)tan(x)

    Auxiliary Equation: mē+1=0 thus m= +/- 1

    CF: Y1= Acos(x)+Bsin(x)

    then I let y=A(x)cos(x)+B(x)sin(x)

    and differentiate to

    y'=-Asin(x)+Bcos(x)+[A'cos(x)+B'sin(x)] and set A'cos(x)+B'sin(x)=0 * (remember this for later)

    differentiating again

    y''= -Acos(x)-Bsin(x)-A'sin(x)+B'cos(x)

    Substituting the yields now:

    -A'sin(x)+B'cos(x)=sec(x)tan(x) ** (remember this for the next step)

    so now I have to solve for A' and/or B'

    I have no idea how to do this and when I checked the answer booklet it explained the next step by:

    * cos(x) - ** sin(x): A'=tanē(x) = 1 - sec(x) A= x - tan(x) + C
    and
    * sin(x) - ** cos(x): B'=tan(x) B= ln[sec(x)] + D

    I fully understand how to get up to the point were it says

    * cos(x) - ** sin(x)

    and I do understand how to integrate tan(x) etc, but I have no idea how I am supposed to get from my two marked (*) differentiations to A'= or B'=

    I hope someone can help me as this is extremely frustrating and I have not been able to find anything that could help me, yet.

    Thank you guys for your time, I appreciate it.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Variation of Parameters

    Quote Originally Posted by MuhTheKuh View Post
    I have a problem understanding a certain step using the Variation of Parameters Method.

    take for example

    y''+y=sec(x)tan(x)

    Auxiliary Equation: mē+1=0 thus m= +/- 1

    CF: Y1= Acos(x)+Bsin(x)

    then I let y=A(x)cos(x)+B(x)sin(x)

    and differentiate to

    y'=-Asin(x)+Bcos(x)+[A'cos(x)+B'sin(x)] and set A'cos(x)+B'sin(x)=0 * (remember this for later)

    differentiating again

    y''= -Acos(x)-Bsin(x)-A'sin(x)+B'cos(x)

    Substituting the yields now:

    -A'sin(x)+B'cos(x)=sec(x)tan(x) ** (remember this for the next step)

    so now I have to solve for A' and/or B'

    I have no idea how to do this and when I checked the answer booklet it explained the next step by:

    * cos(x) - ** sin(x): A'=tanē(x) = 1 - sec(x) A= x - tan(x) + C
    and
    * sin(x) - ** cos(x): B'=tan(x) B= ln[sec(x)] + D

    I fully understand how to get up to the point were it says

    * cos(x) - ** sin(x)

    and I do understand how to integrate tan(x) etc, but I have no idea how I am supposed to get from my two marked (*) differentiations to A'= or B'=

    I hope someone can help me as this is extremely frustrating and I have not been able to find anything that could help me, yet.

    Thank you guys for your time, I appreciate it.
    These are your 2 equations:

    A'\cos x+B'\sin x=0 *

    -A'\sin x+B'\cos x=\sec x\tan x **

    In order to eliminate B, we multiply the * equation by \cos x, the ** equation by \sin x and subtract to get

    A'(\cos^2x+\sin^2x)=-\sec x\tan x\sin x

    A'=-\frac{\sin^2x}{\cos^2x}

    =-\tan^2x

    =1-\sec^2x

    In order to eliminate A, we multiply the * equation by \sin x, the ** equation by \cos x and add to get

    B'(\sin^2x+\cos^2x)=\tan x

    B'=\tan x
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  3. #3
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    Re: Variation of Parameters

    Thanks a lot, but somehow I did not know that -sec(x)tan(x)sin(x) are equal to -sinē(x)/cosē(x)
    or that sec(x)tan(x)cos(x) = tan(x)
    I'd hate to bother you (or anyone else) but is there something like a list where I can see that -sec(x)tan(x)sin(x)= ... etc?
    or how someone would put it if i had to take
    tan(x)cos(x)?

    Thanks a lot!
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Variation of Parameters

    Quote Originally Posted by MuhTheKuh View Post
    Thanks a lot, but somehow I did not know that -sec(x)tan(x)sin(x) are equal to -sinē(x)/cosē(x)
    or that sec(x)tan(x)cos(x) = tan(x)
    I'd hate to bother you (or anyone else) but is there something like a list where I can see that -sec(x)tan(x)sin(x)= ... etc?
    or how someone would put it if i had to take
    tan(x)cos(x)?

    Thanks a lot!
    Just rewrite everything in terms of sines and cosines.

    -\sec x\tan x\sin x=-\frac{1}{\cos x}\frac{\sin x}{\cos x}\sin x=-\frac{\sin^2x}{\cos^2x}

    \sec x\tan x\cos x=\frac{1}{\cos x}\tan x\cos x=\tan x
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