Solve differential equation (x+2)y′+4y=(3x + 6)^-2 lnx

Question
Solve differential equation $$(x+2)y′+4y=(3x + 6)^-2 lnx$$

2021-02-25
$$((x+2)y'+4y)/((x+2))=((3x+6)^-2 lnx)/((x+2))$$
$$y'+(4/(x+2))y= lnx/(9(x+2)^3)$$ (1)
Now, we know that solution of differential equation$$(y'+p(x)y= q(x))$$ is given by
$$y(I.F.)= int q(x)(I.F.)dx$$ (2)
where $$I.F.= e^(int p(x)dx)$$
So, finding integrating factor of equation (1)
$$I.F.= e^(int 4/(x+2)dx)$$
$$= e^(4 int dx/(x+2))$$ $$( :' int dx/(x+a)= ln(x+a)+c)$$
$$= e^(4(ln(x+2)))$$
$$= e^(ln (x+2)^4)$$
$$= (x+2)^4$$
So, solution of given differential equation will be
$$y(x+2)^4= int lnx/(9(x+2)^3) (x+2)^4 dx$$
$$y(x+2)^4= 1//9 int (x+2) ln xdx$$
$$y(x+2)^4= 1/9 [ln x int(x+2)dx- int 1/x [int(x+2)dx]dx]$$
$$= 1/9 [((x+2)^2 lnx)/2 - int 1/x ((x+2)^2)/2 dx]$$
$$= ((x+2)^2 lnx)/18-1/2 int (x^2+4x+4)/x dx$$
$$= ((x+2)^2 lnx)/18-1/2(x^2/2+4x+4lnx)+c$$

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