# Find the General Solution

Find the general solution of the given differential equation (use method of variation of parameter)

4y"+y= 2sec (x/2)   -?< x < ?

Fist we the general solution of the homogeneous equation
$4y''+y=0.$
The charactersitic equation is $4r^2+1$. Hence $r=\pm\frac{1}{2}i$. So
$y_g=c_1 y_1+c_2y_2$
where
$y_1=\sin(\frac{x}{2}), y_2=\cos(\frac{x}{2}).$
The Wronskian is
$W(y_1,y_2)=y_1 y_2'-y_2 y_1'=- \frac{1}{2}\sin^2(\frac{x}{2})-\frac{1}{2}\cos^2(\frac{x}{2})=-\frac{1}{2}.$
By variation of parameters a particular solution is given by
$y_p=-y_1 \int \frac{y_2 (2\sec (\frac{x}{2}))}{W(y_1,y_2)}dx+y_2 \int \frac{y_1 (2\sec (\frac{x}{2}))}{W(y_1,y_2)}dx$
$=4 \sin(\frac{x}{2})\int \cos (\frac{x}{2}) \sec (\frac{x}{2})dx -4 \cos(\frac{x}{2})\int \sin(\frac{x}{2}) \sec (\frac{x}{2})dx$
$=4 \sin (\frac{x}{2}) \int 1 dx - 4\cos (\frac{x}{2}) \int \tan(\frac{x}{2})dx$
$=4 x\sin (\frac{x}{2}) -4 \cos (\frac{x}{2}) (2\ln |\sec \frac{x}{2} |)$
$=4 x\sin (\frac{x}{2}) -8 \cos (\frac{x}{2}) \ln |\sec \frac{x}{2} |.$
So the general solution is
$y=y_g+y_p=c_1 \sin(\frac{x}{2})+c_2\cos(\frac{x}{2})+4 x\sin (\frac{x}{2}) -8 \cos (\frac{x}{2}) \ln |\sec \frac{x}{2} |.$

• Daniel90