# Solve differential equation xy'+[(2x+1)/(x+1)]y= x-1

Solve differential equation$x{y}^{\prime }+\left[\left(2x+1\right)/\left(x+1\right)\right]y=x-1$
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Gennenzip

$xdy/dx+\left(\left(2x+1\right)/\left(x+1\right)\right)y=x-1$
Dividing whole equation by x
$dy/dx+\left(\left(2x+1\right)/\left(x+1\right)\right)y=1/x\left(x-1\right)$
$dy/dx+\left(\left(2x+1\right)/\left(x+1\right)\right)y=1-1/x$
$dy/dx+Py+Q$
$P=\left(2x+1\right)/\left({x}^{2}+x\right)$, $Q=1-1/x$
$I.F.={e}^{\int Pdx}={e}^{\int \left(2x+1\right)/\left({x}^{2}+x\right)dx}$
Let ${x}^{2}+x=t$ then $\left(2x+1\right)dx=dt$
$I.F.={e}^{\int Pdx}={e}^{\int \left(2x+1\right)/\left({x}^{2}+x\right)dx}={e}^{\int 1/tdt}={e}^{\mathrm{log}t}=t={x}^{2}+x$
Then solution is y $I.F.=\int Q\left(I.F.\right)dx+c$
$y\left({x}^{2}+x\right)=\int \left(1-1/x\right)\left({x}^{2}+x\right)dx+c$
$y\left({x}^{2}+x\right)=\int \left({x}^{2}+x-x-1\right)dx+c$
$y\left({x}^{2}+x\right)=\int \left({x}^{2}-1\right)dx+c$
$y\left({x}^{2}+x\right)={x}^{3}/x-x+c$
Multilpying by 3
$3y\left({x}^{2}+x\right)={x}^{3}-3x+3c$
$3y\left({x}^{2}+x\right)={x}^{3}-3x+3{c}^{\prime }$ (where ${c}^{\prime }=3c$)

Jeffrey Jordon