Show that the first order differential equation (x + 1)y'-3y= (x+1)^5 is of the linear type. Hence solve for y given that y = 1.5 when x = 0

$(x+1)y^{\prime }-3y=(x+1{)}^5$
${\displaystyle y^{\prime }-\frac{3}{(x+1)}y={(x+1)}^4}$
Compare with
$y^{\prime }+P(x)y=Q(x)$
${\displaystyle P\left(x\right)=-\frac{3}{x+1}}$
$Q(x)=(x+1{)}^4$
Hence given differential equation is of the linear type
${\displaystyle I.F.=e^{\int P\left(x\right)dx}}$
${\displaystyle =e^{\int -\frac{3}{x+1}dx}}$
${\displaystyle =e^{-3\ln (x+1)}}$
${\displaystyle =e^{\ln {(x+1)}^{-3}}}$
${\displaystyle ={(x+1)}^{-3}}$
${\displaystyle y\cdot I.F.=\int Q\left(x\right)\cdot I.F.dx}$
${\displaystyle y{(x+1)}^3=\int {(x+1)}^4\cdot {(x+1)}^{-3}dx}$
${\displaystyle y{(x+1)}^{-3}=\int (x+1)dx}$
${\displaystyle y{(x+1)}^{-3}=\frac{x^2}{2}+x+C}$
${\displaystyle y=(\frac{x^2}{2}+x+C){(x+1)}^3}$
$y(0)=1.5$
$(0+0+C)(0+1{)}^3=1.5$
C=1.5
${\displaystyle y=(\frac{x^2}{2}+x+1.5){(x+1)}^3}$