Find the general solution of `(dy)/(dx)-3y=sin 2x`

Explanation

Given, `" "(dy)/(dx)-3y=sin2x`
which is a linear differential equation.
On comparing it with `(dy)/(dx)+Py=Q`, we get
`" "P=-3, Q=sin2x`
`" "IF=e^(-3intdx)=e^(-3x)`
The general solution is
`" "y*e^(-3x)=intunderset("I")sin2xunderset("II")e^(-3x)dx`
Let `" "y*e^(-3x)=I" "` ...(i)
`therefore" "I=intunderset("II")(e^(-3x))underset("I")(sin2x)`
`rArr" "I=sin2x(*(e^(-3x))/(-3))-int2cosx((e^(-3x))/(-3))dx+C_(1)`
`rArr" "I=-(1)/(3)e^(-3x)sin2x+(2)/(3)intunderset("II")(e^(-3x))underset("I")(cos2x)dx+C_(1)`
`rArr" "I=-(1)/(3)e^(-3x)sin2x+(2)/(3)(cos2x(e^(-3x))/(-3)-int(-2sin2x)(e^(-3x))/(-3)dx)+C_(1)+C_(2)`
`rArr" "I=-(1)/(3)e^(-3x)sin2x-(2)/(9)cos2xe^(-3x)-(4)/(9)I+C'" "["where,"C'=C_(1)+C_(2)]`
`rArr" "I+(4I)/(9)2=+e^(-3x)(-(1)/(3)sin2x-(2)/(9)cos2x)+C'`
`rArr" "(13)/(9)I=e^(-3x)(-(1)/(3)sin2x-(2)/(9)cos2x)+C'`
`rArr" "I=(9)/(13)e^(-3x)(-(1)/(3)sin2x-(2)/(9)cos2x)+C" "["where"C=(9C')/(13)]`
`rArr" "I=(3)/(13)e^(-3x)(-sin2x-(2)/(3)cos2x)+C`
`rArr" "=(3)/(13)e^(-3x)((-3sin2x-2cos2x))/(3)+C`
`rArr" "=(e^(-3x))/(13)(-3sin2x-2cos2x)+C`
On substituting the value of `I` in Eq. (i), we get
`" "y*e^(-3x)=-(e^(-3x))/(13)(2cos2x+3sin2x)+C`
`rArr" "y=-(1)/(13)(2cos2x=3sin2x)+Ce^(3x)`