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

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

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

Answers (1)

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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