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,

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

Stumbling Robot

Find solutions of y′ – 4y = 2e^xy^1/2 for given initial values

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