Dillard

2021-01-15

Show that the first order differential equation $\left(x+1\right){y}^{\prime }-3y=\left(x+1{\right)}^{5}$ is of the linear type.
Hence solve for y given that y = 1.5 when x = 0

Asma Vang

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

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