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Thread: Checking differential equation (2)

  1. #1
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    Checking differential equation (2)

    Can someone check my working?

    Solve $\displaystyle x^2y''-2xy'+2y=0$ by letting $\displaystyle y=x^m$

    $\displaystyle y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2}$

    $\displaystyle m(m-1)x^2x^{m-2}-2mx^{m-1}x+2x^m=0$

    $\displaystyle m(m-1)x^m-2mx^m+2x^m=0$

    $\displaystyle x^m(m(m-1)-2m+2)=0$

    $\displaystyle x^m(m^2-m-2m+2)=0$

    $\displaystyle x^m(m^2-3m+2)=0$

    $\displaystyle x^m(m-1)(m-2)=0$

    $\displaystyle \Rightarrow m=1, 2$

    $\displaystyle y=x$ or $\displaystyle y=x^2$
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  2. #2
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    Looks good to me.
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