Find the general solution of y′′ + y′ - 2y = ex / (1 + ex)

Find the general solution of the second-order differential equation

$y'' + y' - 2y = \frac{e^x}{1+e^x}.$

If the solution is not valid everywhere, describe the interval on which it is valid.

The general solution of the homogeneous equation

$y'' + y' - 2y = 0$

is given by Theorem 8.7 with $a = 1$ and $b = -2$ . This gives us $d = a^2 - 4b = 9$ ; hence, $k = \frac{1}{2} \sqrt{d} = \frac{3}{2}$ and we have

$\begin{align*} y &= e^{-\frac{ax}{2}} (c_1 e^{kx} + c_2 e^{-kx}) \\ &= e^{-\frac{x}{2}} (c_1 e^{\frac{3}{2}x} + c_2 e^{-\frac{3}{2}x} ) \\ &= c_1 e^x + c_2 e^{-2x}. \end{align*}$

So, we obtain particular solutions of the of the homogeneous equation $v_1(x) = e^x$ and $v_2(x) = e^{-2x}$ (by taking $c_1 = 1, c_2 = 0$ and $c_1 = 0, c_2 = 1$ , respectively). We want to apply Theorem 8.9 (on page 330 of Apostol). From that theorem we have

$R(x) = \frac{e^x}{1+e^x}.$

Furthermore,

$\begin{align*} W(x) &= v_1 v_2' - v_1' v_2 \\ &= e^x (-2e^{-2x}) - e^x e^{-2x} \\ &= -3 e^{-x}. \end{align*}$

So, a particular solution $y_1$ to the non-homogeneous equation is given by

$\begin{align*} y_1(x) &= t_1 (x) v_1(x) + t_2(x) v_2(x) \\ t_1(x) &= -\int v_2(x) \frac{R(x)}{W(x)} \, dx \\ &= -\int \left( \frac{1}{e^{2x}} \right) \left( \frac{e^x}{1+e^x} \right) \left( \frac{1}{-3e^{-x}}\right) \, dx \\ &= \frac{1}{3} \int \frac{1}{1+e^x} \, dx \\ &= \frac{1}{3} (x - \log (e^x + 1)) \\ t_2(x) &= \int v_1(x) \frac{R(x)}{W(x)} \, dx \\ &= \int e^x \left( \frac{e^x}{1+e^x} \right) \left( \frac{-e^x}{3} \right) \\ &= -\frac{1}{3} \int \left( \frac{e^{3x}}{1+e^x} \right) \, dx \\ &= -\frac{1}{3} \left(e^{-2x} \log (e^x + 1) + \frac{1}{2} e^{2x} - e^x \right). \end{align*}$

So, we have a particular solution of the non-homogeneous equation given by

$\begin{align*} y_1 &= \frac{1}{3} e^x \left( x - \log (e^x + 1) \right) - \frac{1}{3e^{2x}} \left( \frac{1}{e^{2x}} \log (e^x + 1) + \frac{1}{2} e^{2x} - e^x \right) \\ &= \frac{1}{3} xe^x - \frac{1}{3} e^x \log (e^x + 1) - \frac{1}{3} e^{-4x} \log (e^x + 1) - \frac{1}{6} + \frac{1}{3} e^{-x}. \end{align*}$

Therefore, the general solution of the non-homogeneous equation is

$y = e^x \left( c_1 + \frac{1}{3} x \right) + c_2 e^{-2x} - \frac{1}{3} \log (e^x + 1) (e^x + e^{-2x}) + \frac{1}{3} e^{-x} - \frac{1}{6}.$

Stumbling Robot

Find the general solution of y′′ + y′ – 2y = e^x / (1 + e^x)

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