Question

Using the method of vibration of parameters,solve the equation
d^2y/dx^2 +a^2y =secax

Answer 1

$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + a^2y = \sec(ax) \\ \textsf{The solution to the above equation is} \\ y = y_c + y_p \\ \textsf{where}\, y_c \, \textsf{is the complementary factor and} \\ y_p\, \textsf{ is the particular integral} \\ \textsf{The auxiliary equation is} \,m^2 + a^2 = 0 \\ \begin{aligned} (m - ja)(m + ja) &= 0 \\ \therefore m &= \pm ja \end{aligned}\\ \{\textsf{where}\, j\, \textsf{is a complex number}\}\\ \textsf{Recall that if the solution of the auxiliary equation}\\ \textsf{of a second-order differential equation is}\\ m = \alpha \pm j\beta, \textsf{the general solution is} \\ y = e^{\alpha x}(C_1\cos{\beta x} + C_2\sin{\beta x})\\ \therefore y_c = C_1\cos(ax) + C_2\sin(ax)\\ \textsf{The Wronskian of the two solutions is} \\ W(x) = \begin{vmatrix} \cos(ax) & \sin(ax)\\ \frac{\mathrm{d}}{\mathrm{d}x}(\cos(ax)) & \frac{\mathrm{d}}{\mathrm{d}x}(\sin(ax)) \end{vmatrix} = \begin{vmatrix} \cos(ax) & \sin(ax)\\ -a\sin(ax) & a\cos(ax) \end{vmatrix} \\ \begin{aligned} &= a\cos^2(ax) + a\sin^2(ax)\\ &= a(\cos^2(ax) + \sin^2(ax)) \\ &= a \end{aligned}\\ \therefore\textsf{Our particular solution will be given by}\\ y_p = V_1(x)\cos(ax) + V_2(x)\sin(ax)\\ \textsf{Where}\, V_1(x) = -\int \frac{r(x)\sin(ax)}{W(x)}\, \mathrm{d}x \, \textsf{and} \, V_2(x) = \int \frac{r(x)\cos(ax)}{W(x)} \, \mathrm{d}x\\ \begin{aligned} V_1(x) = -\int \frac{r(x)\sin(ax)}{W(x)} \, \mathrm{d}x \\ &= -\int \frac{\sec(ax)\sin(ax)}{a}\, \mathrm{d}x\\ &= -\int \frac{\tan(ax)}{a}\, \mathrm{d}x\\ &= -\frac{\ln(\sec(ax))}{a^2} + C \end{aligned} \\ \begin{aligned} V_2(x) &= \int \frac{r(x)\cos(ax)}{W(x)}\, \mathrm{d}x \\ &= \int \frac{\sec(ax)\cos(ax)}{a}\, \mathrm{d}x\\ &= \int \frac{1}{a}\, \mathrm{d}x\\ &= \frac{x}{a} + C \end{aligned} \\ \textsf{The constant terms of the integration can be} \\\textsf{ignored since we are trying to find a non-constant} \\\textsf{solution to the differential equation} \\ \begin{aligned} \therefore y &= y_c + y_p \\ &=C_1\cos(ax) + C_2\sin(ax) + \frac{x}{a}\sin(ax) - \frac{1}{a^2}\ln(\sec(ax))\cos(ax) \end{aligned}$