Answer to Question #295405 in Differential Equations for Moni

Question

Solve by the method of variation of parameter y''+y=secx

Answer 1

The corresponding homogeneous differential equation is

$$y''+y=0$$

Characteristic (auxiliary) equation

$$r^2+1=0$$

$$r=\pm i$$

The general solution of the homogeneous differential equation is

$$y_h=c_1 \cos x+c_2\sin x$$

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

If

$$c_1 '\cos x+c_2'\sin x=0,$$

then

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

Substitute

$$-c_1'\sin x-c_1\cos x+c_2'\cos x-c_2\sin x$$

$$+c_1 \cos x+c_2\sin x=\dfrac{1}{\cos x}$$

We have

$$c_1 '\cos x+c_2'\sin x=0,$$ $$-c_1'\sin x+c_2'\cos x=\dfrac{1}{\cos x}$$

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

$$\dfrac{\sin^2 x}{\cos x}c_2'+c_2'\cos x=\dfrac{1}{\cos x}$$

$$c_1'=-\dfrac{\sin x}{\cos x}$$

$$c_2'=1$$

Integrate

$$c_1=-\int \dfrac{\sin x}{\cos x}dx=\ln (|\cos x|)+C_1$$

$$c_2=\int dx=x+C_2$$

The general solution of the nonhomogeneous differential equation is

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