Question

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

Answer 1

\((x+1)y'−3y=(x+1)^5\)
\(\displaystyle{y}'-\frac{3}{{{\left({x}+{1}\right)}}}{y}={\left({x}+{1}\right)}^{4}\)
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
\(\displaystyle{y}={\left(\frac{{x}^{2}}{{2}}+{x}+{1.5}\right)}{\left({x}+{1}\right)}^{3}\)