Answer to Question #297970 in Differential Equations for praveen

Question

solve by method of variation of parameter y'' + y = sec x

Answer 1

Solution:

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 non-homogeneous differential equation is

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