Results 1 to 7 of 7

Math Help - Linear Differential Equation:

  1. #1
    Member kjchauhan's Avatar
    Joined
    Nov 2009
    Posts
    137
    Thanks
    2

    Linear Differential Equation:

    Please solve this...

    x^2\frac{d^2 y}{dx} - 2x(1+x)\frac{dy}{dx}+2(1+x)y=x^2

    I need solution of this equation..
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2009
    Posts
    31
    If I remember correctly; you first calculate the homogeneous solution. And then the particular solution for the non-homogeneous part.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Deviding both terms by x^{2} the DE becomes...

    y^{''} - 2\cdot \frac{1+x}{x}\cdot y^{'} + 2\cdot \frac{1+x}{x^{2}}\cdot y = 1 (1)

    First step is the solution of 'incomplete equation' ...


    y^{''} - 2\cdot \frac{1+x}{x}\cdot y^{'} + 2\cdot \frac{1+x}{x^{2}}\cdot y = 0 (2)

    ... that is of the type...

    y^{''} + p(x) \cdot y^{'} + q(x)\cdot y = 0 (3)

    A DE like (3) can often be solved in terms of an auxiliary function u(x) defined as...

    \ln y = \ln u -\frac{1}{2}\cdot \int p(x)\cdot dx (4)

    Deriving both terms of (4) we obtain first ...

    \frac {y^{'}}{y} = \frac{u^{'}}{u} - \frac{1}{2}\cdot p(x) (5)

    ... deriving again...

    \frac{y\cdot y^{''}}{y^{2}} - (\frac{y^{'}}{y})^{2}= \frac{u\cdot u^{''}}{u^{2}} - (\frac{u^{'}}{u})^{2} - \frac{1}{2}\cdot p^{'} (x) (6)

    ... and then with two more steps...

    \frac{y^{''}}{y}= \frac{u^{''}}{u}+\frac{1}{4}\cdot p^{2}(x) -\frac{1}{2}\cdot p^{‘}(x) (7)

    Taking into account (7) the (3) can be written in terms of u(x) as...

    u^{''} + \theta(x)\cdot u =0 (8)

    ... where...

    \theta(x)= q(x) - \frac{1}{4}\cdot p^{2}(x) -\frac{1}{2}\cdot p^{'}(x) (9)

    A little pause before proceeding...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 19th 2009 at 05:58 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    In the last post we have 'attacked' the 'incomplete' DE...

    y^{''} - 2\cdot \frac{1+x}{x}\cdot y^{'} + 2\cdot \frac{1+x}{x^{2}}\cdot y = 0 (1)

    ... swapping the y with the u variable defined as...

    \ln y = \ln u -\frac{1}{2}\cdot \int p(x)\cdot dx (2)


    ... and we are arrived to the DE...

    u^{''} + \theta(x)\cdot u= 0 (3)

    ... where...

    \theta (x)= q(x) - \frac{1}{4}\cdot p^{2} (x) - \frac{1}{2}\cdot p^{'} (x) (4)

    In our case we have...

    p(x)= -2\cdot (1+\frac{1}{x}), q(x)= \frac{2}{x}\cdot (1+\frac{1}{x}) (5)

    ... so that...

    y= u\cdot x\cdot e^{x} ,  \theta(x)=-1 (6)

    The (6) shows us that the solution of (1) in terms of u is relatively 'easy' and is...

     u(x)= c_{1}\cdot e^{-x}+ c_{2}\cdot e^{x} (7)

    ... so that is...

     y(x)= c_{1}\cdot x+c_{2}\cdot x\cdot e^{2x} (8)

    The solution of the 'complete' DE is left for further study...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 19th 2009 at 07:10 AM. Reason: Corrected the mistake found by showsend...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Hi Chisigma. I get \theta(x)=-1. Or no?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,391
    Thanks
    55
    Quote Originally Posted by kjchauhan View Post
    Please solve this...

    x^2\frac{d^2 y}{dx} - 2x(1+x)\frac{dy}{dx}+2(1+x)y=x^2

    I need solution of this equation..
    Just a note. Sometimes a clever substitution works very well. Your ODE can be written as

    x^2\frac{d^2 y}{dx} - 2(1+x)\left( x \frac{dy}{dx} - y\right)=x^2.

    If you let

     <br />
u = x \frac{dy}{dx} - y\;\;\;(1)<br />
then your entire problem reduces to

    x \frac{du}{dx} - 2(x+1)u = x^2 a linear ODE! Once you solve this then go to (1) and solve again a linear ODE.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Continuing with Chisigma's solution, then I'll finish it if that's alright. I'll use variation of parameter:

    Then u(x)=c_1 e^{x}+c_2 e^{-x} and therefore y(x)=\left(c_1 e^{x}+c_2 e^{-x}\right) \exp\left(\int_x (1/z+1)dz\right)=\left(c_1 e^{x}+c_2 e^{-x}\right) xe^{x}=c_1 x e^{2x}+c_2 x

    Let y(x)=A(x) xe^{2x}+B(x) x

    Substituting that into the original DE we obtain the following two equations to solve:

    A'xe^{2x}+B'x=0

    A'(2xe^{2x}+e^{2x})+B'=x^2

    Then B'=-A'e^{2x}

    Substituting that one into the second one, I get A'=1/2x e^{-2x} or B'=-1/2 x so that B(x)=-1/4 x^2 and A(x)=-1/8 e^{-2x}-1/4 x e^{-2x} in which we drop the arbitrary constants in this case since we're only looking for a particular solution to add to the complimentary solution.

    Therefore, the general solution is y(x)=c_1 xe^{2x}+c_2 x-1/4 x^3+xe^{2x}\left(-1/8 e^{-2x}-1/4 x e^{-2x}\right)

    or:

    y(x)=c_1 x e^{2x}+c_2 x-1/4 x^3-1/4 x^2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with non linear differential equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: November 20th 2010, 06:40 AM
  2. Matlab-Linear Differential Equation
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 22nd 2010, 12:12 PM
  3. Linear Differential Equation:
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 10th 2009, 12:17 PM
  4. non linear differential equation help please
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 30th 2009, 06:43 AM
  5. linear differential equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 21st 2009, 07:20 AM

Search Tags


/mathhelpforum @mathhelpforum