Answer to Question #279080 in Differential Equations for kalashnikov

Question

(D^2 + 4) y = 4 sec 2x csc 2x

Answer 1

The corresponding homogeneous equation is

$$(D^2+4)y=0$$

Auxiliary equation

$$r^2+4=0$$

$$r_1=2i, r_2=-2i$$

The general solution of the homogeneous equation is

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

Variation of Parameters

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

Let

$$c_1'\cos(2x)+c_2'\sin(2x)=0$$

Then

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

$$y''=-2c_1'\sin(2x)-4c_1\cos(2x)$$

$$+2c_2'\cos(2x)-4c_2\sin(2x)$$

Substitute

$$-2c_1'\sin(2x)-4c_1\cos(2x)$$

$$+2c_2'\cos(2x)-4c_2\sin(2x)$$

$$+4c_1\cos(2x)+4c_2\sin(2x)$$

$$=4\sec(2x)\csc(2x)$$

$$\begin{cases} c_1'\cos(2x)+c_2'\sin(2x)=0\\ \\ -2c_1'\sin(2x)+2c_2'\cos(2x)=4\sec(2x)\csc(2x) \end{cases}$$

$$c_2'=-\dfrac{\cos(2x)}{\sin(2x)}c_1'$$

$$-c_1'\sin(2x)-\dfrac{\cos^2(2x)}{\sin(2x)}c_1'=\dfrac{2}{\sin(2x)\cos(2x)}$$

$$c_1'=-\dfrac{2}{\cos(2x)}$$

Integrate

$$c_1=-\int \dfrac{2}{\cos(2x)}dx$$

$$\int \dfrac{2}{\cos(2x)}dx=\int \dfrac{2\cos(2x)}{\cos^2(2x)}dx$$

$$=\int \dfrac{2\cos(2x)}{1-\sin^2(2x)}dx$$

$$=\int \dfrac{\cos(2x)}{1-\sin(2x)}dx+\int \dfrac{\cos(2x)}{1+\sin(2x)}dx$$

$$=-\dfrac{1}{2}\int \dfrac{d(1-\sin(2x))}{1-\sin(2x)}dx+\dfrac{1}{2}\int \dfrac{d(1+\sin(2x))}{1+\sin(2x)}dx$$

$$=-\dfrac{1}{2}\ln(|1-\sin(2x)|)+\dfrac{1}{2}\ln(|1+\sin(2x)|)+C_3$$

$$=\dfrac{1}{2}\ln\dfrac{(1+\sin(2x))^2}{1-\sin^2(2x)}+C_3$$

$$=\ln(1+\sin(2x))-\ln(|\cos(2x)|)+C_3$$

$$c_1=-\ln(1+\sin(2x))+\ln(|\cos(2x)|)-C_3$$

$$c_2'=\dfrac{2}{\sin(2x)}$$

Integrate

$$c_2=\int \dfrac{2}{\sin(2x)}dx$$

$$\int \dfrac{2}{\sin(2x)}dx=\int \dfrac{2\sin(2x)}{\sin^2(2x)}dx$$

$$=\int \dfrac{2\sin(2x)}{1-\cos^2(2x)}dx$$

$$=\int \dfrac{\sin(2x)}{1-\cos(2x)}dx+\int \dfrac{\sin(2x)}{1+\cos(2x)}dx$$

$$=\dfrac{1}{2}\int \dfrac{d(1-\cos(2x))}{1-\cos(2x)}dx-\dfrac{1}{2}\int \dfrac{d(1+\cos(2x))}{1+\cos(2x)}dx$$

$$=\dfrac{1}{2}\ln(|1-\cos(2x)|)-\dfrac{1}{2}\ln(|1+\cos(2x)|)+C_4$$

$$=-\dfrac{1}{2}\ln\dfrac{(1+\cos(2x))^2}{1-\cos^2(2x)}+C_4$$

$$=-\ln(1+\cos(2x))+\ln(|\sin(2x)|)+C_4$$

$$c_2=-\ln(1+\cos(2x))+\ln(|\sin(2x)|)+C_4$$

The general solution of the homogeneous equation is

$$y_h=c_5\cos(2x)+c_6\sin(2x)$$

$$+\bigg(-\ln(1+\sin(2x))+\ln(|\cos(2x)|)\bigg)\cos(2x)$$

$$+\bigg(-\ln(1+\cos(2x))+\ln(|\sin(2x)|)\bigg)\sin(2x)$$