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Thread: 2nd order differential equation problem

  1. #1
    Member nautica17's Avatar
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    2nd order differential equation problem

    I just wanted to confirm I did this right.

    "Verify by substitution that the given function is a solution of the given differential equation."

    Given:
    x^2 y'' - xy' + 2y = 0
    y = x*cos( ln(x) )

    My work:
    y' = cos(ln(x)) - sin(ln(x))
    y'' = -(1/x) * ( sin(ln(x)) + cos(ln(x)) )

    Substituting back in to the differential equation:
    x^2 (-(1/x)(sin(ln(x)) + cos(ln(x)))) - x(cos(ln(x)) - sin(ln(x))) - 2(xcos(ln(x)) = 0

    -xsin(ln(x)) - xcos(ln(x)) - xcos(ln(x)) + xsin(ln(x)) + 2x*cos(ln(x)) = 0

    0 = 0

    Does this mean I verified the solution? See I'm confused because in first order diff. equations I would just separate out the x's and y's and integrate both sides and get my y equation. This is a little different apparently.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: 2nd order differential equation problem

    Quote Originally Posted by nautica17 View Post
    Does this mean I verified the solution?
    Yes, you have proved that the given function is a solution of the differential equation.

    See I'm confused because in first order diff. equations I would just separate out the x's and y's and integrate both sides and get my y equation. This is a little different apparently.
    The order of the equation is irrelevant. One thing is to solve a differential equation and another thing is to verify that a given function is a solution.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: 2nd order differential equation problem

    Quote Originally Posted by FernandoRevilla View Post
    Yes, you have proved that the given function is a solution of the differential equation.



    The order of the equation is irrelevant. One thing is to solve a differential equation and another thing is to verify that a given function is a solution.
    The DE...

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

    ... is 'Euler's type' and a particular solution of it has the form $\displaystyle x^{\alpha}$. Substituting $\displaystyle x^{\alpha}$ in (1) You arrive at the algebraic equation...

    $\displaystyle \alpha^{2}-2\ \alpha +2 =0$ (2)

    ...the solution of which are $\displaystyle 1 \pm i$, so that the general solution of (1) is...

    $\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (3)

    Of course if You want a real solution it must be $\displaystyle c_{1}=c_{2}=c$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: 2nd order differential equation problem

    Quote Originally Posted by chisigma View Post
    The DE...

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

    ... is 'Euler's type' and a particular solution of it has the form $\displaystyle x^{\alpha}$. Substituting $\displaystyle x^{\alpha}$ in (1) You arrive at the algebraic equation...

    $\displaystyle \alpha^{2}-2\ \alpha +2 =0$ (2)

    ...the solution of which are $\displaystyle 1 \pm i$, so that the general solution of (1) is...

    $\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (3)

    Of course if You want a real solution it must be $\displaystyle c_{1}=c_{2}=c$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    This confirms that it is not the same to find the general solution of a DE that verify that a given function is a solution of a DE.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: 2nd order differential equation problem

    Quote Originally Posted by chisigma View Post
    The DE...

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

    ... is 'Euler's type' and a particular solution of it has the form $\displaystyle x^{\alpha}$. Substituting $\displaystyle x^{\alpha}$ in (1) You arrive at the algebraic equation...

    $\displaystyle \alpha^{2}-2\ \alpha +2 =0$ (2)

    ...the solution of which are $\displaystyle 1 \pm i$, so that the general solution of (1) is...

    $\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (3)

    Of course if You want a real solution it must be $\displaystyle c_{1}=c_{2}=c$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    My answer has been a little hurried so that what i wrote is not precise... the general solution of the De is...

    $\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (1)

    ... bue $\displaystyle c_{1}$ and $\displaystyle c_{2}$ may be complex, so that the general real solution is...

    $\displaystyle y(x)= x\ (a_{1}\ \cos \ln x + a_{2}\ \sin \ln x)$

    ... where $\displaystyle a_{1}$ and $\displaystyle a_{2}$ are 'arbitrary real constants'...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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