Answer to Question #345991 in Differential Equations for john

Question

1. Find the general solution of the differential equation using the method

of variation of parameters y′′ + 4y = 4sec(2x)

Answer 1

The corresponding homogeneous differential equation is

$$y''+4y=0$$

Corresponding (auxiliary) equation is

$$r^2+4=0$$

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

The general solution of the homogeneous differential equation is

$$y_h=C_1\cos(2x)+C_2\sin(2x)$$

Consider

$$y_1=C_1\cos(2x)+C_2\sin(2x)$$

$$y_1'=C_1'\cos(2x)-2C_1\sin(2x)+$$

$$+C_2'\sin(2x)+2C_2\cos(2x)$$

Let $C_1'\cos(2x)+C_2'\sin(2x)=0.$ Then

$$y_1'=-2C_1\sin(2x)+2C_2\cos(2x)$$

$$y_1''=-2C_1'\sin(2x)-4C_1\cos(2x)$$

$$+2C_2'\cos(2x)-4C_2\sin(2x)$$

Substitute

$$-2C_1'\sin(2x)-4C_1\cos(2x)$$

$$+2C_2'\cos(2x)-4C_2\sin(2x)$$

$$+4C_1\cos(2x)+4C_2\sin(2x)=4\sec (2x)$$

Solve the system

$$C_1'\cos(2x)+C_2'\sin(2x)=0$$ $$-2C_1'\sin(2x)+2C_2'\cos(2x)=\dfrac{4}{\cos(2x)}$$

$$C_1'\cos(2x)+C_2'\sin(2x)=0$$ $$-C_1'\sin(2x)\cos(2x)+C_2'\cos^2(2x)=2$$

$$C_1'\cos(2x)+C_2'\sin(2x)=0$$ $$C_2'=2$$

$$C_1'=-2\dfrac{\sin(2x)}{\cos(2x)}$$ $$C_2'=2$$

Integrate

$$C_1=-2\int \dfrac{\sin(2x)}{\cos(2x)}dx=\ln|\cos(2x)|+C_3$$

$$C_2=2x+C_4$$

The general solution of the non homogeneous differential equation is

$$y=\ln|\cos(2x)|\cos(2x)+C_3\cos(2x)$$ $$+2x\sin(2x)+C_4\sin(2x)$$