Priscilla Johnston
2022-01-19
Answered

For the sulotion of the OED: $\frac{dy}{dx}=1+x+{y}^{2}+xy$ .

You can still ask an expert for help

yotaniwc

Answered 2022-01-19
Author has **34** answers

Changing $y=u-\frac{z}{2}$ the equation is cast to the form

$\frac{du}{dx}={u}^{2}+{(\frac{x}{2}-1)}^{2}+\frac{5}{2}$

which does not seem to possess solutions in terms of elementary functions

which does not seem to possess solutions in terms of elementary functions

limacarp4

Answered 2022-01-20
Author has **39** answers

if we know $y}_{p}=-\frac{{x}^{2}-2x-1}{{x}^{2}-4x+3$ is a particular solution. So we can find general solution of Riccati Equation. (Note:I did not try it .I trust GEdgar. Thanks to him. I just offer a general solution If we know a particular solution)

Use transform of$y={y}_{p}+\frac{1}{H}$

$\frac{dy}{dx}=1+x+{y}^{2}+xy$ .

$y}_{p}^{{}^{\prime}}+\left(\frac{-H}{{H}^{2}}\right)=1+x+({y}_{p}^{2}+\frac{2{y}_{p}}{H}+\frac{1}{{H}^{2}})+x{y}_{p}+\frac{x}{H$

$y}_{p}^{{}^{\prime}}=1+x+{y}_{p}^{2}+x{y}_{p}+\frac{{H}^{\prime}}{{H}^{2}}+\frac{2{y}_{p}}{H}+\frac{1}{{H}^{2}}+\frac{x}{H$

if$y}_{p$ is a solution of the equation then it must satisfy

$y}_{p}^{{}^{\prime}}=1+x+{y}_{p}^{2}+x{y}_{p$

Thus$0=\frac{{H}^{\prime}}{{H}^{2}}+\frac{2{y}_{p}}{H}+\frac{1}{{H}^{2}}+\frac{x}{H}$

Use transform of

if

Thus

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thanks

$\frac{dy}{dx}}=f(x,y)$

where $f(x,y)$ is a homogeneous function. I found some examples like $f(x,y)=(x+y{)}^{2}$ where it can be solved after converting it to Ricatti's equation.

thanks

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