Solve ` x (dy)/(dx) + 2y = x cosx `

Explanation

The given differential eqution may be written as
` (dy)/(dx) + (2)/(x) * y = cos x " "` ... (i)
This is of the form ` (dy)/(dx) +Py =Q`, where `P = (2)/(x) and Q = cos x `
Thus, the given equation is linear.
`IF = e ^(int P dx ) = e ^( int (2)/(x) dx) = e ^(2 log x ) = e ^( log (x ^(2)) ) = x ^(2)`
So, the required solution is
`y xx IF = int {Qxx (IF)}dx + C`,
i.e., `yx ^(2) = int x ^(2) cos x dx + C `
` = x ^(2) sinx - int 2x sin x dx + C ` [integrating by parts]
` " " = x ^(2) sinx - 2 * [ x(-cos x ) - int 1* (-cos x ) dx ] + C ` [ integrating by parts]
` " " = x ^(2) sin x + 2x cos x - 2 sinx +C `
Hence, `y = sin x + (2)/(x) cos x - (2)/(x ^(2)) sinx + (C)/( x ^(2)) ` is the required solution.