Find the particular solution of the differential equation. `(dy)/(dx)+ycotx=4x\ \ cos e c\ x ,\ (x!=0),\ ` given that `y=0` when `x=pi/2dot`

Explanation

Given differential equation is :
`(dy)/(dx)+ycotx=4x cosecx`……`(1)`
Comparing with the differential equation `(dy)/(dx)+Py=Q`,
`P=cotx`, `Q=4x cosecx` and `I.F.=e^(intPdx)`
`:. I.F.=e^(cotxdx)impliesI.F.=e^(log|sinx|)=sinx`
Therefore, the general solution of the given differential equation is
`y*I.F=intQxxI.F.dx+c`
`implies y sinx=int4x cosecx*sinxdx+c`
`implies ysinx=int4xdx+Cimplies ysinx=4*(x^(2))/(2)+C`
`implies ysinx=2x^(2)+C`.......`(2)`
Now, `x=(pi)/(2)`if `y=0`
`:. 0xxsin"(pi)/(2)=2((pi)/(2))^(2)+CimpliesC=-(pi^(2))/(2)`
Put the value of `C` in equation `(2)`,
`y sinx=2x^(2)-(pi^(2))/(2)`