Find the general solution of the equation y′′ + 9y = sin (3x)

For a nonzero constant $k$ prove that

$y_1 = \frac{1}{k} \int_0^x R(t) \sin k(x-t) \, dt,$

is a particular solution of the differential equation

$y'' + k^2 y = R(x).$

Find the general solution of the equation

$y'' + 9y = \sin 3x.$

Proof. The general solution of $y'' + k^2 y = 0$ is $y = c_1 \cos (kx) + c_2 \sin (kx)$ . So, we have particular solutions given by $v_1 = \cos (kx)$ and $v_2 = \sin (kx)$ (taking $c_1 = 1, \ c_2 = 0$ and $c_1 = 0, \ c_2 = 1$ , respectively). The Wronskian of $v_1$ and $v_2$ is

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

So, we have the functions $t_1$ and $t_2$ of Theorem 8.9 given by

$\begin{align*} t_1 (x) &= -\int v_2 \frac{R}{W} \, dx \\ &= -\frac{1}{k} \int_0^x \sin (kt) R(t) \, dt \\ t_2 (x) &= \int v_1 \frac{R}{W} \, dx \\ &= \frac{1}{k} \int_0^x \cos(kt) R(t) \, dt. \end{align*}$

This means we have a particular solution $y_1$ of the non-homogeneous equation given by

$\begin{align*} y_1 &= t_1 v_1 + t_2 v_2 \\ &= -\frac{\cos (kx)}{k} \int_0^x \sin (kt) R(t) \, dt + \frac{\sin(kx)}{k} \int_0^x \cos(kt) R(t) \, dt\\ &= \frac{1}{k} \left( \int_0^x R(t) \big( \cos(kt) \sin (kx) - \cos (kx) \sin (kt) \big) \, dt \right) \\ &= \frac{1}{k} \int_0^x R(t) \sin (k(x-t)) \, dt. \qquad \blacksquare \end{align*}$

Using this theorem we can compute the general solution of

$y'' + 9y = \sin (3x).$

We have

$\begin{align*} y &= c_1 \cos (3x) + c_2 \sin (3x) + \frac{1}{3} \int_0^x (\sin (3t))\sin (3 (x-t)) \, dt \\ &= c_1 \cos (3x) + c_2 \sin (3x) + \frac{-x}{6} \cos(3x) - \frac{1}{18} \sin (3x) \\ &= \left(c_1 - \frac{x}{6} \right) \cos (3x) + \left( c_2 - \frac{1}{18} \right) \sin (3x). \end{align*}$

Stumbling Robot

Find the general solution of the equation y′′ + 9y = sin (3x)

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