Results 1 to 5 of 5

Thread: Non-Homogeneous Second grade diff. equation to solve

  1. #1
    Newbie Black Kawairothlite's Avatar
    Joined
    Oct 2008
    Posts
    21

    Non-Homogeneous Second grade diff. equation to solve

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

    Consider the equation

    $\displaystyle y''+y=f(x)$

    where

    $\displaystyle y(0)=X_0 ; y'(0)=V_0$

    prove that the solution is:

    $\displaystyle y(x)=\int_{0}^{x}sin(x-s)f(s)ds + X_0cosx+V_0senx$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2008
    Posts
    120
    I just tried using Laplace transforms but i get a slightly different solution. Almost the same. What method did you use?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027
    You are NOT asked to solve the equation. You are given a function and asked to show it is the solution.

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

    You may want to use
    $\displaystyle \frac{d }{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt= F(x, \beta(x))\frac{d\beta}{dx}- F(x,\alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial F(x,t)}{\partial x} dt$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie Black Kawairothlite's Avatar
    Joined
    Oct 2008
    Posts
    21
    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 $\displaystyle y(x)= \int_0^x sin(s-x)f(s)ds+ X_0 cos(x)+ V_0 sin(x)$
    what is y'(x)? what is y"(x)? What is y(0)? What is y'(0)?

    You may want to use
    $\displaystyle \frac{d }{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt= F(x, \beta(x))\frac{d\beta}{dx}- F(x,\alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial F(x,t)}{\partial x} dt$
    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........
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie Black Kawairothlite's Avatar
    Joined
    Oct 2008
    Posts
    21
    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...


    $\displaystyle y''+y=f(x)$

    so i got the general solution

    $\displaystyle y''+y=0$

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

    then the general solution is:

    $\displaystyle y_g=c_1sen x + c_2cos x$

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

    so:

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

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

    where $\displaystyle y_1=sen x ; y_2=cos x$

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

    then

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

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

    so

    $\displaystyle v_1=\int f(x)cos x dx$

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

    and the particular solution is:

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

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

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

    and finally

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

    now the initial values:

    $\displaystyle y(0)=X_0 ; y'(0)=V_0$

    $\displaystyle y(0)=c_1sen(0)+c_2cos(0)=X_0$ so $\displaystyle c_2=X_0$

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

    then the solution for the initial values is:

    $\displaystyle y(x)=\int_{0}^{s} (cos ssen x ds-cos xsen s)f(s) ds + X_0cos x + V_0 sen x$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: May 3rd 2011, 11:39 AM
  2. Replies: 0
    Last Post: May 3rd 2011, 11:28 AM
  3. 1st order homogeneous diff. equation
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: Jan 31st 2010, 10:36 AM
  4. 2nd Order homogeneous diff equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Jan 28th 2010, 11:43 AM
  5. Cant solve this diff equation :X
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Oct 13th 2009, 11:55 AM

Search Tags


/mathhelpforum @mathhelpforum