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Find solutions of y′ – 4y = 2exy1/2 for given initial values

Find all solutions of the nonlinear differential equation

    \[ y' - 4y = 2e^x y^{\frac{1}{2}} \]

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

    \[ y = 2 \qquad \text{when} \qquad x = 0. \]


From the 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

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

In the present problem we have the equation

    \[ y' - 4y = 2e^x y^{\frac{1}{2}} \qquad \text{on } (-\infty, +\infty) \qquad \text{with } y = 2 \text{ when } x = 0. \]

Therefore, to apply the previous exercise we have

    \[ P(x) = -4, \quad Q(x) = 2e^x, n = \frac{1}{2}, k = 1 - n = \frac{1}{2}. \]

Hence, f(x) is a solution to the given equation if and only if (f(x))^{\frac{1}{2}} = g(x) where g(x) is the unique solution to

    \[ v' - 2 v = e^x \qquad \text{with } g(0) = \sqrt{2}. \]

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

    \[ A(x) = \int_0^x -2 \,dt = -2x, \]

giving us

    \begin{align*}  g(x) &= \sqrt{2} e^{2x} + e^{2x} \int_0^x 2e^t \cdot e^{-2t} \, dt \\  &= \sqrt{2} e^{2x} + e^{2x} (-e^{-x} + 1) \\  &= \sqrt{2} e^{2x} + e^{2x} - e^x. \end{align*}

Therefore,

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

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