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

Solve differential equation $\left(x+2\right)y\prime +4y=\left(3x+6{\right)}^{-}2\mathrm{ln}x$

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