Find the General Solution

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

4y"+y= 2sec (x/2)   -π< x < π


Fist we the general solution of the homogeneous equation 
The charactersitic equation is $4r^2+1$. Hence $r=\pm\frac{1}{2}i$. So 
\[y_g=c_1 y_1+c_2y_2\]
\[y_1=\sin(\frac{x}{2}),  y_2=\cos(\frac{x}{2}). \]
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
\[y=y_g+y_p=c_1 \sin(\frac{x}{2})+c_2\cos(\frac{x}{2})+4 x\sin (\frac{x}{2}) -8 \cos (\frac{x}{2}) \ln |\sec \frac{x}{2} |.\]

  • If possible, please set a later deadline for your future posts. Ideally at least 24 hours.

The answer is accepted.
Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.