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Thread: Second Order Linear Differential Equation

  1. #1
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    Second Order Linear Differential Equation

    If anyone can guide me through this problem, it would be much helpful. Thank you.

    Let u be a nonzero solution of the second-order equation

    y'' + P(x)y' + Q(x)y = 0

    Show that the substitution y = uv converts the equation

    y'' + P(x)y' + Q(x)y = R(x)

    into a first-order linear equation for v'
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  2. #2
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    Quote Originally Posted by hashshashin715 View Post
    If anyone can guide me through this problem, it would be much helpful. Thank you.

    Let u be a nonzero solution of the second-order equation

    y'' + P(x)y' + Q(x)y = 0

    Show that the substitution y = uv converts the equation

    y'' + P(x)y' + Q(x)y = R(x)

    into a first-order linear equation for v'
    let

    $\displaystyle y=uv$
    then
    $\displaystyle y'=u \frac{dv}{dx}+\frac{du}{dx}v$
    and
    $\displaystyle y''=u \frac{d^2v}{dx^2}+2\frac{du}{dx}\frac{dv}{dx}+\fra c{d^2u}{dx^2}v$
    Substitute these values in the given equation and determine u
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  3. #3
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    Ok, but if I do that, I will still have u' and u'' left over. What do I do with them? Aren't I only suppose to express the equation in terms of v's and their derivatives only?
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  4. #4
    Member kjchauhan's Avatar
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    Quote Originally Posted by hashshashin715 View Post
    Ok, but if I do that, I will still have u' and u'' left over. What do I do with them? Aren't I only suppose to express the equation in terms of v's and their derivatives only?

    After Substituting, we get,

    $\displaystyle (u''v+2u'v'+uv'')+P(u'v+uv')+Quv=R$

    $\displaystyle \therefore uv''+(ru'+Pu)v'+(u''+Pu'+Qu)v=R$

    Since $\displaystyle u$is a nonzero solution, then,

    $\displaystyle u''+Pu'+Qu=0$

    $\displaystyle \therefore uv''+(2u'+Pu)v'=R$

    $\displaystyle \therefore v''+\left(\frac{2}{u}u'+P\right)v'=\frac{R}{u}$

    Now put$\displaystyle v'=p$, so that, $\displaystyle v''=\frac{dp}{dx}$, then we have,

    $\displaystyle \frac{dp}{dx}+\left(\frac{2}{u}u'+P\right)p=\frac{ R}{u}$

    Which is a linear equation in p,

    $\displaystyle \therefore$ I.F.=$\displaystyle e^{\int \left(\frac{2}{u}u'+P\right)dx}$ $\displaystyle =u^2e^{\int P dx}$

    $\displaystyle \therefore pu^2e^{\int P dx}=\int \left(\frac{R}{u}u^2e^{\int P dx}\right)dx+c_1$
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  5. #5
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    Oh, I see. Thank you
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