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Math Help - Non-Homogeneous Second grade diff. equation to solve

  1. #1
    Newbie Black Kawairothlite's Avatar
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    Non-Homogeneous Second grade diff. equation to solve

    Hey all, i gonna share another problem to solve here we go:

    Consider the equation

    y''+y=f(x)

    where

    y(0)=X_0 ; y'(0)=V_0

    prove that the solution is:

    y(x)=\int_{0}^{x}sin(x-s)f(s)ds + X_0cosx+V_0senx
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  2. #2
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    I just tried using Laplace transforms but i get a slightly different solution. Almost the same. What method did you use?
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  3. #3
    MHF Contributor

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    You are NOT asked to solve the equation. You are given a function and asked to show it is the solution.

    If y(x)= \int_0^x sin(s-x)f(s)ds+ X_0 cos(x)+ V_0 sin(x)
    what is y'(x)? what is y"(x)? What is y(0)? What is y'(0)?

    You may want to use
    \frac{d }{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt= F(x, \beta(x))\frac{d\beta}{dx}- F(x,\alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial F(x,t)}{\partial x} dt
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  4. #4
    Newbie Black Kawairothlite's Avatar
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    Quote Originally Posted by whipflip15 View Post
    I just tried using Laplace transforms but i get a slightly different solution. Almost the same. What method did you use?
    characteristic equation of second grade ordinary diff eq. and then parameter variation method... that's the way i think is done

    Quote Originally Posted by HallsofIvy View Post
    You are NOT asked to solve the equation. You are given a function and asked to show it is the solution.

    If y(x)= \int_0^x sin(s-x)f(s)ds+ X_0 cos(x)+ V_0 sin(x)
    what is y'(x)? what is y"(x)? What is y(0)? What is y'(0)?

    You may want to use
    \frac{d }{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt= F(x, \beta(x))\frac{d\beta}{dx}- F(x,\alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial F(x,t)}{\partial x} dt
    that's cheating lol i know it'd work but u gotta get the solution by solving the diff eq... what if you didn't have the answer........
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  5. #5
    Newbie Black Kawairothlite's Avatar
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    well i got the answer few days ago, been busy lately but for the record im gonna post the answer..as usual if im wrong correct me plz...


    y''+y=f(x)

    so i got the general solution

    y''+y=0

    so m^2+1=0
    m=\pm(i) remember that i=\sqrt-1

    then the general solution is:

    y_g=c_1sen x + c_2cos x

    now i gotta get the particular solution which i got with parameters variation method...

    so:

    y_p=v_1(x)sen x+ v_2(x)cos x where v_1(x) and v_2(x) are unknown but according to the formula:

    v'_1=\frac{{-y_2f(x)}}{{W(y_1,y_2)}} ; v'_2=\frac{{y_1f(x)}}{{W(y_1,y_2)}}

    where y_1=sen x ; y_2=cos x

    and W=y_1y'_2-y_2y'_1=-(sen^2 x+cos^2 x)=-1

    then

    v'_1=\frac{{-cos xf(x)}}{{-1}}

    v'_2=\frac{{y_1sen xf(x)}}{{-1}}

    so

    v_1=\int f(x)cos x dx

    v_2=-\int f(x)sen x dx

    and the particular solution is:

    y_p=sen x\int f(x)cos x dx-cos x\int f(x)sen x dx

    y_p=\int_{0}^{s} cos sf(s)sen x ds-\int_{0}^{s} cos xf(s)sen s ds

    y_p=\int_{0}^{s} (cos ssen x ds-cos xsen s)f(s) ds

    and finally

    y_p=\int_{0}^{s} sen (x-s)f(s) ds

    now the initial values:

    y(0)=X_0 ; y'(0)=V_0

    y(0)=c_1sen(0)+c_2cos(0)=X_0 so c_2=X_0

    y'(0)=c_1cos(0)-c_2sen(0)=V_0 so c_1=V_0

    then the solution for the initial values is:

    y(x)=\int_{0}^{s} (cos ssen x ds-cos xsen s)f(s) ds + X_0cos x + V_0 sen x
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