# Define the term homogeneous first order ordinary differential equation. Hence solve x^2 y' + xy = x2 + y

Rui Baldwin 2021-02-09 Answered
Define the term homogeneous first order ordinary differential equation.
Hence solve ${x}^{2}{y}^{\prime }+xy=x2+y$
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Maciej Morrow

$dy/dx=F\left(y/x\right)$
Now solve the given differential equation
${x}^{2}{y}^{\prime }+xy={x}^{2}+y$
${y}^{\prime }+\left(xy\right)/{x}^{2}=\left({x}^{2}+y\right)/{x}^{2}=>{y}^{\prime }+y/x=1+y/{x}^{2}$
${y}^{\prime }+y/x-y/{x}^{2}=1=>{y}^{\prime }+y\left(1/x-1/{x}^{2}\right)=1$
$P=1/x-1/{x}^{2},Q=1$
$IF={e}^{\int Pdx}={e}^{\int \left(1/x-1/{x}^{2}\right)dx}$
$={e}^{\mathrm{log}x+1/x}={e}^{\mathrm{log}x}{e}^{1/x}$
$IF=x{e}^{1/x}$
Thus, the solution of differential equation is
$y\left(IF\right)=\int IF\cdot Q\left(x\right)dx+C$
Solving further to get
$y\left(x{e}^{1/x}\right)=\int x{e}^{1/x}\cdot 1dx+C$
$y\left(x{e}^{1/x}\right)=x\int {e}^{1/x}dx-\int 1\int {e}^{1/x}dx+C$
$y\left(x{e}^{1/x}\right)=\left(x{e}^{1/x}\right)/\left(-1/{x}^{2}\right)+\int {e}^{1/x}/\left(-1/{x}^{2}\right)dx+C$
$⇒y\left(x{e}^{1/x}\right)={x}^{3}{e}^{1/x}+\int {x}^{2}{e}^{1/x}dx+C$
$=-{x}^{3}{e}^{1/x}+{x}^{2}\int {e}^{1/x}dx+\int 2x\int {e}^{1/x}dx+C$
$=-{x}^{3}{e}^{1/x}+1/3{e}^{1/x}{x}^{3}+1/6{e}^{1/x}{x}^{2}+1/6{e}^{1/x}x-1/6{E}_{i}\left(1/x\right)+C$
${E}_{i}\left(x\right)$ is the exponential integral
$y\left(x{e}^{1/x}\right)=1/2{e}^{1/x}{x}^{2}-1/2\left(-{e}^{1/x}x+{E}_{i}\left(1/x\right)\right)+C$
$y\left(x\right)=1/2x-1/2\left(-1+\left({E}_{i}\left(1/x\right)\right)/\left(x{e}^{1/x}\right)\right)+C/\left(x{e}^{1/x}\right)$

Jeffrey Jordon