Determine the general solution of y′′ + y = sin x

Determine the general solution of the second-order differential equation

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

The homogeneous equation related to this is an equation of the form $y'' + ay' + by = 0$ with $a = 0$ and $b = 1$ . This gives us $d = a^2 - 4b = -4$ and $k = \frac{1}{2} \sqrt{-d} = 1$ . So, the general solution of the homogeneous equation $y'' + y = 0$ is given by

$y = c_1 \cos x + c_2 \sin x.$

Then by the previous exercise we know that a particular solution $y_1$ for the non-homogeneous equation is given by

$\begin{align*} y_1 &= \frac{1}{k} \int_0^x R(t) \sin (k(x-t)) \, dt\\ &= \int_0^x (\sin t)(\sin (x-t)) \, dt \\ &= \int_0^x (\sin t)(\sin x \cos t - \sin t \cos x) \, dt \\ &= -\cos x \int_0^x \sin^2 t \, dt + \sin x \int_0^x \frac{\sin (2t)}{2} \, dt\\ &= \frac{1}{2} \sin^3 x - \frac{1}{2} x \cos x + \frac{1}{2} \sin x \cos^2 x. \end{align*}$

Therefore, the general solution of the given equation is

$\begin{align*} y &= \left( c_1 - \frac{x}{2} \right) \cos x + \left( c_2 + \frac{1}{2} (\sin^2 x + \cos^2 x) \right) \sin x \\ &= \left( c_1 - \frac{x}{2} \right) \cos x + c_2 \sin x. \end{align*}$

(Where we absorbed the $\frac{1}{2}$ in the value of the constant $c_2$ .)

Stumbling Robot

Determine the general solution of y′′ + y = sin x

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply