# 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

First order differential equations
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

2021-01-16

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