Answer to Question #102573 in Differential Equations for Ajay

Question

Apply the method of variations of parameters to solve the following differential equations: y" +a^2y=cosec ax

Answer 1

$y''+a^2y=\cosec ax\\ y''+a^2y=\frac{1}{\sin ax}\\$

Solution can be found in the form

$y=y_1+y_2$

1) $y_1$ is solution of equation

$y''+a^2y=0\\ \lambda^2+a^2=0\\ \lambda_1=ia, \lambda_2=-ia,\\ f_1=\cos ax, f_2=\sin ax\\$

then

$y_1=c_1 \cos ax+c_2 \sin ax$

2) solution y can be found by the method of variation of constants in the form

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

functions $c_1(x)$ and $c_2(x)$ can be found from the system of equations

$\left\{ \begin{matrix} c'_1(x) \cos ax+c'_2(x) \sin ax =0\\ c'_1(x)(- a\sin ax)+c'_2(x) a \cos ax =\frac{1}{\sin ax} \end{matrix} \right.$

$\Delta=\left| \begin{matrix} \cos ax & \sin ax\\ -a \sin ax & a\cos ax \end{matrix} \right|=\\ =a\cos^2 ax+a \sin^2 ax=a\\ \Delta_1=\left| \begin{matrix} 0 & \sin ax\\ \frac{1}{\sin ax} & a\cos ax \end{matrix} \right|=-1\\ \Delta_2=\left| \begin{matrix} \cos ax &0\\ -a \sin ax & \frac{1}{\sin ax} \end{matrix} \right|= \frac{\cos ax}{\sin ax}\\ c'_1(x)=\frac{\Delta_1}{\Delta}=\frac{-1}{a}, \\ c'_2(x)=\frac{\Delta_2}{\Delta}=\frac{1}{a} \frac{\cos ax}{\sin ax}, \\ c_1(x)=\int \frac{-1}{a} dx=\frac{-x}{a} + k_1\\ c_2(x)=\int\frac{1}{a} \frac{\cos ax}{\sin ax}dx=\\ =\frac{1}{a^2} \int\frac{d(\sin ax)}{\sin ax}=\frac{1}{a^2}\ln|\sin ax|+k_2\\ k_1, k_2 \, are\,constants$

Then

$y=\left(\frac{-x}{a} + k_1\right)\cos ax + \left(\frac{1}{a^2}\ln|\sin ax|+k_2\right)\sin ax$