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

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