Determine the general solution of y′′ + 4y = 3x cos x

Determine the general solution of the second-order differential equation

$y'' + 4y = 3x \cos x.$

The homogeneous equation related to this is an equation of the form $y'' + ay' + by = 0$ with $a = 0$ and $b = 4$ . This gives us $d = a^2 - 4b = -16$ and $k = \frac{1}{2} \sqrt{-d} = 2$ . So, the general solution of the homogeneous equation $y'' + 4y = 0$ is given by

$y = c_1 \cos (2x) + c_2 \sin (2x).$

Then by a previous exercise (Section 8.17, Exercise #19) we know that a particular solution $y_1$ for the non-homogeneous equation is given by

$\begin{align*} y_1 &= \frac{1}{k} \int_0^x R(t) \sin (k(x-t)) \, dt\\ &= \frac{1}{2} \int_0^x 3t \cos t \sin (2(x-t)) \, dt \\ &= \frac{1}{4}x \cos x - \frac{1}{12} \sin x + \frac{3}{4} x \cos x + \frac{3}{4} \sin x - \frac{1}{12} \sin (2x) + 3 \sin (2x). \end{align*}$

Therefore, the general solution of the given equation is

$y = c_1 \cos (2x) + c_2 \sin (2x) + x \cos x + \frac{2}{3} \sin x.$

Stumbling Robot

Determine the general solution of y′′ + 4y = 3x cos x

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