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Find all solutions of the differential equation (x – 2) (x – 3)y′ + 2y = (x – 1) (x – 2)

Find all solutions of the differential equation

    \[ (x-2)(x-3) y' + 2y = (x-1)(x-2) \]

on the following intervals:

  1. (-\infty, 2);
  2. (2,3);
  3. (3, +\infty).

Prove that all of the solutions have a finite limit as x \to 2, and that none of the solutions have a finite limit as x \to 3.


First, some general setup that will apply to all parts. Since on any of the three interval we have (x-2)(x-3) \neq 0 we can divide through by this term to obtain,

    \[ (x-2)(x-3) y' + 2y = (x-1)(x-2) \quad \implies \quad y' + \frac{2}{(x-2)(x-3)} y = \frac{x-1}{x-3}. \]

We will then apply Theorem 8.3 (page 310 of Apostol) with

    \[ P(x) = \frac{2}{(x-2)(x-3)}, \qquad Q(x) = \frac{x-1}{x-3}. \]

In each part will choose a different value for a and give the solutions in terms b.

  1. On the interval (-\infty, 2) we choose the initial value a = 1. Then we have (using partial fractions to evaluate the integral)

        \[ A(x) = \int_1^x \frac{2}{(t-2)(t-3)} \, dt = 2 \left( \log |3-x|  - \log |2-x| - \log 2 \right). \]

    Therefore,

        \begin{align*}  f(x) &= \frac{b e^{2 \log|2-x| + 2 \log 2}}{e^{2 \log |3-x|}} + \frac{e^{2 \log|2-x| + 2 \log 2}}{e^{2 \log|3-x|}} \int_1^x \frac{t-1}{t-3} \frac{e^{2 \log|3-t|}}{e^{2 \log |2-t| + 2 \log 2}} \, dt \\[9pt]  &= b \frac{4 (x-2)^2}{(x-3)^2} + \frac{4(x-2)^2}{(x-3)^2} \int_1^x \frac{(t-1)(t-3)}{4(t-2)^2} \, dt \\[9pt]  &= \frac{4 (x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} \int_1^x \frac{t^2 - 4t + 3}{t^2 - 4t + 4} \, dt \right) \\[9pt]  &= \frac{4 (x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} \left( \int_1^x \, dt - \int_1^x \frac{1}{(t-2)^2} \, dt \right) \right)\\[9pt]  &= \frac{4 (x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} \left( x-1 + \frac{1}{t-2} \Bigr \rvert_1^x \right) \right)\\[9pt]  &= \frac{4(x-2)^2}{(x-3)^2} \left( b + \frac{1}{4} x + \frac{1}{4(x-2)} \right)\\[9pt]  &= \left( \frac{x-2}{x-3} \right)^2 \left( x + \frac{1}{x-2} + C \right), \end{align*}

    where C is the constant \frac{b}{4}.

  2. On the interval (2,3) the same solution is going to work, with the different initial condition getting absorbed in the constant term. We can go ahead and compute (borrowing from our computations in part (a) where appropriate). Choose a = \frac{5}{2} \in (2,3) giving us

        \begin{align*}   A(x) &= \int_{\frac{5}{2}}^x \frac{2}{(t-2)(t-3)} \, dt \\  &= 2 \left( \log |3-x| - \log |2-x| - \log \left|3 - \frac{5}{2}\right| + \log \left|2 - \frac{5}{2}\right| \right) \\  &= 2 (\log|3-x| - \log|2-x|). \end{align*}

    Therefore,

        \begin{align*}  f(x) &= \frac{b e^{2 \log |2-x|}}{e^{2 \log|3-x|}} + \frac{e^{2 \log |2-x|}}{e^{2 \log |3-x|}} \int_{\frac{5}{2}}^x \frac{t-1}{t-3} \frac{e^{2 \log |3-t|}}{e^{2 \log |2-t|}} \, dt \\[9pt]  &= b \frac{(x-2)^2}{(x-3)^2} + \frac{(x-2)^2}{(x-3)^2} \int_{\frac{5}{2}}^x \frac{(t-1)(t-3)}{(t-2)^2} \, dt \\[9pt]  &= \frac{(x-2)^2}{(x-3)^2} \left( b + \int_{\frac{5}{2}}^x \frac{t^2 - 4t + 3}{t^2 - 4t + 4} \, dt \right) \\[9pt]  &= \frac{(x-2)^2}{(x-3)^2} \left( b + (x - 1) + \frac{1}{t-2}\Bigr \rvert_{\frac{5}{2}}^x \right) \\[9pt]  &= \frac{(x-2)^2}{(x-3)^2} \left( b + x - 1 + \frac{1}{x-2} + 2 \right) \\[9pt]  &= \left(\frac{(x-2)}{(x-3)}\right)^2 \left( x + \frac{1}{x-2} + C \right), \end{align*}

    where C is the constant b +1.

  3. This will follow identically to parts (a) and (b) arriving at the same result. The different initial condition getting absorbed in the constant C.

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