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Find solutions of 2xyy′ + (1+x)y2 = ex for given initial values

Find all solutions of the nonlinear differential equation

    \[ 2xyy' + (1+x)y^2 = e^x \]

on the interval (0, +\infty) satisfying the initial conditions

  1. y = \sqrt{e} when x = 1;
  2. y = -\sqrt{e} when x = 1;
  3. a finite limit as x \to 0.

    From a previous exercise (Section 8.5, Exercise #13) we know that a function y = f(x) which is never zero on an interval I is a solution of the initial value problem

        \[ y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } (f(a))^k = b \]

    if and only if (f(x))^k = g(x) on I where g(x) = v is the unique solution of the initial-value problem

        \[ v' + kP(x) v = kQ(x) \qquad \text{on } I, \qquad \text{with } g(a) = b. \]

    1. In the present problem we have the equation

          \[ y' + \frac{1+x}{2x} y = \frac{e^x}{2x} y^{-1} \qquad \text{on } (0, +\infty) \qquad \text{with } y = \sqrt{e} \text{ when } x = 1. \]

      Therefore, to apply the previous exercise we have

          \[ P(x) = \frac{1+x}{2x}, \quad Q(x) = \frac{e^x}{2x}, n = -1,  k = 1 - n = 2. \]

      Hence, f(x) is a solution to the given equation if and only if (f(x))^2 = g(x) where g(x) is the unique solution to

          \[ v' + \frac{1+x}{x}v = \frac{e^x}{x} \qquad \text{with } g(1) = e. \]

      We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

          \[ A(x) = \int_1^x \frac{1+t}{t} \,dt = \log x + x - 1, \]

      giving us

          \begin{align*}  g(x) &= e \cdot e^{-\log x - x +1} + e^{-\log x - x +1} \int_1^x \frac{e^t}{t} e^{\log t + t - 1} \, dt \\[9pt]  &= \frac{e^2}{xe^x} + \frac{e}{xe^x} \int_1^x e^{2t-1} \, dt \\[9pt]  &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \int_1^x e^{2t} \, dt\\[9pt]  &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \left( \frac{1}{2} e^{2t} \Bigr \rvert_1^x \right) \\[9pt]  &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \left( \frac{e^{2x}}{2} - \frac{e^2}{2} \right) \\[9pt]  &= \frac{e^x + e^2 - x}{2x} \end{align*}

      Therefore,

          \[ (f(x))^2 = g(x) \quad \implies \quad f(x) = \left( \frac{e^x+e^{2-x}}{2x} \right)^{\frac{1}{2}}. \]

    2. From part (a) we have g(x) = (f(x))^2 which implies f(x) = \pm (g(x))^{\frac{1}{2}}. Since f(1) = -\sqrt{e} we then have

          \[ f(x) = - \left( \frac{e^x + e^{2-x}}{2x} \right)^{\frac{1}{2}}. \]

    3. Let \lim_{x \to 0} f(x) = b where b is some finite real number. Then, (using the calculations we already completed in part (a) and just changing the initial-values a and b)

          \begin{align*}  v &= be^{-x-\log x + a + \log a} + e^{-x - \log x + a + \log a} \int_a^x \let( \frac{e^t}{t} \right) \left( e^{t + \log t - a - \log a} \right) \, dt \\[9pt]  &=\frac{bae^a}{xe^x} + \left( \frac{1}{xe^x} \right) \int_a^x e^{2t} \, dt \\[9pt]  &= \frac{bae^a}{xe^x} + \frac{e^{2x} - e^{2a}}{2xe^x} \\[9pt]  &= \frac{e^{2x} - c}{2xe^x}. \end{align*}

      Then,

          \[ \lim_{x \to 0} \frac{e^{2x} - c}{2xe^x} = L \quad \implies \quad \lim_{x \to 0} (e^{2x} - c) = 0 \quad \implies \quadd c = 1. \]

      Therefore,

          \[ y^2 = \frac{e^{2x}-1}{2xe^x} = \frac{e^x - e^{-x}}{2x} = \frac{\sinh x}{x}. \]

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