How to solve the differential equation $y'=\frac{y-xy^2}{x+x^2y}$

Question

$$y'=\frac{y-xy^2}{x+x^2y}$$

This is the equation I want to solve.

My idea is to substitute $xy$ with $u,$ $u=xy$ and $\frac{du}{dx}=xy'+y.$

So, the equation becomes

$y'=\frac{y(1-u)}{x(1+u)}.$

What I am struggling with is how to deal with the $x$ and $y$ that are left in the equation.

Answer 1

The equation is $$y^{\prime}=\frac{y-x y^2}{x+x^2 y}$$

HINT: We can re-write it as: $$-\frac{1}{y^2}\frac{dy}{dx}=\frac{\left(x-\frac{1}{y}\right)}{x+x^2 y}$$

Now letting $\frac{1}{y}=t \Rightarrow -\frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$ $$\Rightarrow \frac{d t}{d x}=\frac{t(x-t)}{x(x+t)}=\frac{\frac{t}{x}\left(1-\frac{t}{x}\right)}{1+\frac{t}{x}}$$

Now use $t=vx \Rightarrow \frac{dt}{dx}=v+x\frac{dv}{dx}$

$$\Rightarrow \quad v+x \frac{d v}{d x}=\frac{v(1-v)}{1+v} \Rightarrow x \frac{d v}{d x}=\frac{-2 v^2}{1+v}$$

Answer 2

Starting from your last line: $$y'=\frac{y(1-u)}{x(1+u)}.$$ Then: $$xy'=\frac{y(1-u)}{(1+u)}.$$ $$u'-y=\frac{y(1-u)}{(1+u)}.$$ $$u'=\frac{y(1-u)+y(1+u)}{(1+u)}.$$ $$u'=\frac{2y}{(1+u)}$$ $$xu'=\frac{2xy}{(1+u)}$$ $$xu'=\frac{2u}{(1+u)}$$ $$x\dfrac {du}{dx}=\frac{2u}{(1+u)}$$ The DE is separable:

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Another solution:

$$y'=\frac{y-xy^2}{x+x^2y}$$ $$y'x-y=-xy^2-x^2yy'$$ $$d\dfrac {y}{x}=-\dfrac 12 \dfrac {d(x^2y^2)}{x^2}$$ $$\dfrac x yd\dfrac {y}{x}=- {d(xy)}$$ Integrate. $$\ln \dfrac yx+xy=C$$

Answer 3

$(xy^{2}-y)~dx+(yx^{2}+x)~dy=0$

Integrating factor :$μ(x,y)=x^{m} y^{n}$

$(x^{m}y^{n+2}-x^{m}y^{n+1})~dx+(y^{n+1}x^{m+2}+x^{m+1}y^{n})~dy=0$

\begin{cases} \frac{\partial M}{\partial y}=(n+2)x^{m}y^{n+1}-(n+1)x^{m}y^{n} \\\\ \frac{\partial N}{\partial x}=(m+2)x^{m+1}y^{n+1}+(m+1)x^{m}y^{n}\\\\ \end{cases}

The condition of exact of the differential equation :$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

\begin{cases} m+2=n+2 \\\\ -n-1=m+1\\\\ \end{cases}

$m=n=-1$

$μ(x,y)=x^{-1} y^{-1}$

$(y-x^{-1})~dx+(x+y^{-1})~dy=0$

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=1$

$f(x,y)=\int(y-x^{-1})~dx+h(y)=xy-\ln(x)+h(y)$

$x+h'(y)=x+y^{-1}$

$h(y)=\int y^{-1}~dy=\ln(y)$

$f(x,y)=xy-\ln(x)+\ln(y)=c$

$$xy+\ln(\frac{y}{x})=C$$