Alissa Hancock

2022-07-16

How can I solve this first order linear differential equation?
${y}^{\prime }=1-\frac{2}{x+y}$
I have tried turning it into an inexact differential equation, but I get an integration factor $\mu \left(x,y\right)$ and I don't know how to apply it.

Gornil2

See $v=\frac{2}{x+y},$, then $2\frac{\phantom{\rule{thinmathspace}{0ex}}dv}{\phantom{\rule{thinmathspace}{0ex}}dx}=-{v}^{2}\left(1+\frac{\phantom{\rule{thinmathspace}{0ex}}dy}{\phantom{\rule{thinmathspace}{0ex}}dx}\right)$
The given equation reduces to
$-2\frac{\phantom{\rule{thinmathspace}{0ex}}dv}{{v}^{2}\left(2-v\right)}=\phantom{\rule{thinmathspace}{0ex}}dx.$
Integrate. Use partial fractions for the left hand side. Can you take it from here?

aggierabz2006zw

You can use a substitution to settle this differential equation. Let s(x)=x+y(x). Differentiating this gives us$\frac{ds}{dx}\left(x\right)=1+\frac{dy}{dx}\left(x\right)=1+1-\frac{2}{s\left(x\right)}.$
Algebraic manipulation gives
$\frac{ds}{dx}\left(x\right)=2-\frac{2}{s\left(x\right)}=2\left(1-\frac{1}{s\left(x\right)}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\frac{ds}{dx}\left(x\right)}{1-\frac{1}{s\left(x\right)}}=2.$
Integrating both sides with respect to x gives us
$\int \frac{\frac{ds}{dx}\left(x\right)}{1-\frac{1}{s\left(x\right)}}dx=\int 2dx\phantom{\rule{0ex}{0ex}}⇒\mathrm{ln}\left(s\left(x\right)-1\right)+s\left(x\right)=2x+C$
for a constant C. Rearrange for s(x) and substitute back.

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