Find the general solution of y′′ + y = cot2 x

Find the general solution of the second-order differential equation

$y'' + y = \cot^2 x.$

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

The general solution of the homogeneous equation

$y'' + y = 0$

is given by Theorem 8.7 with $a = 0$ and $b = 1$ . This gives us $d = a^2 - 4b = -4$ ; hence, $k = \frac{1}{2} \sqrt{-d} = 1$ and we have

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

So, we obtain particular solutions of the of the homogeneous equation $v_1(x) = \cos x$ and $v_2(x) = sin x$ (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

$R(x) = \cot^2 x.$

Furthermore,

$\begin{align*} W(x) &= v_1 v_2' - v_1' v_2 \\ &= (\cos x)(\cos x) - (-\sin x)(\sin x) \\ &= \cos^2 x + \sin^2 x \\ &= 1. \end{align*}$

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

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

This implies,

$\begin{align*} y_1 &= \cos x ( \log | \cot x + \csc x| - \cos x) + \sin x \left( \frac{-\sin^2 x - 1}{\sin x} \right) \\ &= \cos x (\log | \cot x + \csc x |) - 2. \end{align*}$

Therefore, the general solution is

$y = \cos x (c_1 + \log | cot x + \csc x|) + c_2 \sin x - 2.$

Stumbling Robot

Find the general solution of y′′ + y = cot² x

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