If `y+d/(dx)(x y)=x(sinx+logx),fin dy(x)dot`

Explanation

Given differential equation is
`" "y+(d)/(dx)(xy)=x(sinx+logx)`
`rArr" "y+x(dy)/(dx)+y=x(sinx+logx)`
`rArr " "x(dy)/(dx)+2y=x(sinx+logx)`
`rArr" "(dy)/(dx)+(2)/(x)y=sinx+logx`
which is a linear differential equation.
On comparing it with `" "(dy)/(dx)+Py=Q,` we get
`" "P=(2)/(x),Q=sinx+logx`
`" "IF=e^(int(2)/(x)dx)=e^(2logx)=x^(2)`
The general solution is
`" "y*x^(2)=int(sinx+logx)x^(2)dx+C`
`rArr" "y*x^(2)=int(x^(2)sinx+x^(2)logx)dx+C`
`rArr" "y*x^(2)=intx^(2)sinxdx+int x^(2)logxdx+C`
`rArr" "y*x^(2)=I_(1)+I_(2)+C" "`...(i)
Now, `" "I_(1)=intx^(2)sinxdx`
`" "=x^(2)(-cosx)+int2xcosxdx`
`" "=-x^(2)cosx+[2x(sinx)-int2sinxdx]`
`" "I_(1)=-x^(2)cosx+2xsinx+2cosx" "`...(ii)
and `" "I_(2)=intx^(2)logxdx`
`" "=logx*(x^(3))/(3)-int(1)/(x)*(x^(3))/(3)dx`
`" "=logx*(x^(3))/(3)-(1)/(3)intx^(2)dx`
`" "=logx*(x^(3))/(3)-(1)/(3)*(x^(3))/(3)" "`...(iii)
On substituting the value of `I_(1) and I_(2)` in Eq. (i), we get
`" "y*x^()=-x^(2)cosx+2xsinx+2cosx+(x^(3))/(3)logx-(1)/(9)x^(3) +C`
`therefore " "y=-cosx+(2sinx)/(x)+(2cosx)/(x)+(x)/(3)logx-(x)/(9)+Cx^(-2)`