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Find the solution of the differential equation xy′′ – y′ + (1 – x)y = 0

Assume the differential equation

    \[ xy'' - y' + (1-x)y = 0 \]

has a solution of the form

    \[ y = e^{mx} \]

for some constant m. Determine an explicit formula for this solution.


Incomplete.

One comment

  1. Evangelos says:

    Since we are given

        \begin{align*} y &= e ^{mx} \end{align*}

    We then know that

        \begin{align*} y' &= me ^{mx} \\ y'' &= m^2 e^{mx} \end{align*}

    And we have

        \begin{align*} xy'' - y' + (1-x)y &= x(m^2 e^{mx}) - me^{mx} + (1-x)e^{mx} \\ &= 0 \end{align*}

    Since e^mx can never be equal to zero, we can then divide both sides by it

        \begin{align*} m^{2} x - m + (1-x) &= (m^{2} - 1)x + (1 - m) \\ &= 0  \end{align*}

    But that would mean that
    m^{2} – 1 & = 0 \\
    1 – m &= 0
    \end{align*}

    Thus, m = 1, and we have

        \begin{align*} y &= e ^{mx} \\ &= e^x \end{align*}

    As requested.

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