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Thread: Linear Differential Equation:

  1. #1
    Member kjchauhan's Avatar
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    Linear Differential Equation:

    Please solve this...

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

    I need solution of this equation..
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  2. #2
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    If I remember correctly; you first calculate the homogeneous solution. And then the particular solution for the non-homogeneous part.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Deviding both terms by $\displaystyle x^{2}$ the DE becomes...

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

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


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

    ... that is of the type...

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

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

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

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

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

    ... deriving again...

    $\displaystyle \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...

    $\displaystyle \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 $\displaystyle u(x)$ as...

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

    ... where...

    $\displaystyle \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

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Nov 19th 2009 at 04:58 AM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    In the last post we have 'attacked' the 'incomplete' DE...

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

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

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


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

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

    ... where...

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

    In our case we have...

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

    ... so that...

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

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

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

    ... so that is...

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Nov 19th 2009 at 06:10 AM. Reason: Corrected the mistake found by showsend...
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  5. #5
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    Hi Chisigma. I get $\displaystyle \theta(x)=-1$. Or no?
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  6. #6
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    Quote Originally Posted by kjchauhan View Post
    Please solve this...

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

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

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

    If you let

    $\displaystyle
    u = x \frac{dy}{dx} - y\;\;\;(1)
    $ then your entire problem reduces to

    $\displaystyle x \frac{du}{dx} - 2(x+1)u = x^2$ a linear ODE! Once you solve this then go to (1) and solve again a linear ODE.
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  7. #7
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    Continuing with Chisigma's solution, then I'll finish it if that's alright. I'll use variation of parameter:

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

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

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

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

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

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

    Substituting that one into the second one, I get $\displaystyle A'=1/2x e^{-2x}$ or $\displaystyle B'=-1/2 x$ so that $\displaystyle B(x)=-1/4 x^2$ and $\displaystyle A(x)=-1/8 e^{-2x}-1/4 x e^{-2x}$ 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 $\displaystyle y(x)=c_1 xe^{2x}+c_2 x-1/4 x^3+xe^{2x}\left(-1/8 e^{-2x}-1/4 x e^{-2x}\right)$

    or:

    $\displaystyle y(x)=c_1 x e^{2x}+c_2 x-1/4 x^3-1/4 x^2$
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