\((x+1)y'−3y=(x+1)^5\)

\(\displaystyle{y}'-\frac{3}{{{\left({x}+{1}\right)}}}{y}={\left({x}+{1}\right)}^{4}\)

Compare with

\(y'+P(x)y= Q(x)\)

\(\displaystyle{P}{\left({x}\right)}=-\frac{3}{{{x}+{1}}}\)

\(Q(x)= (x+1)^4\)

Hence given differential equation is of the linear type

\(\displaystyle{I}.{F}.={e}^{{\int{P}{\left({x}\right)}{\left.{d}{x}\right.}}}\)

\(\displaystyle={e}^{{\int-\frac{3}{{{x}+{1}}}{\left.{d}{x}\right.}}}\)

\(\displaystyle={e}^{{-{3} \ln{{\left({x}+{1}\right)}}}}\)

\(\displaystyle={e}^{{ \ln{{\left({x}+{1}\right)}}^{ -{{3}}}}}\)

\(\displaystyle={\left({x}+{1}\right)}^{ -{{3}}}\)

\(\displaystyle{y}\cdot{I}.{F}.=\int{Q}{\left({x}\right)}\cdot{I}.{F}.{\left.{d}{x}\right.}\)

\(\displaystyle{y}{\left({x}+{1}\right)}^{3}=\int{\left({x}+{1}\right)}^{4}\cdot{\left({x}+{1}\right)}^{ -{{3}}}{\left.{d}{x}\right.}\)

\(\displaystyle{y}{\left({x}+{1}\right)}^{ -{{3}}}=\int{\left({x}+{1}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle{y}{\left({x}+{1}\right)}^{ -{{3}}}=\frac{{x}^{2}}{{2}}+{x}+{C}\)

\(\displaystyle{y}={\left(\frac{{x}^{2}}{{2}}+{x}+{C}\right)}{\left({x}+{1}\right)}^{3}\)

\(y(0)=1.5\)

\((0+0+C)(0+1)^3=1.5\)

C=1.5

\(\displaystyle{y}={\left(\frac{{x}^{2}}{{2}}+{x}+{1.5}\right)}{\left({x}+{1}\right)}^{3}\)