#### 2.101   ODE No. 101

$x y'(x)+x y(x)^2-y(x)=0$ Mathematica : cpu = 0.0707184 (sec), leaf count = 18

$\left \{\left \{y(x)\to \frac {2 x}{2 c_1+x^2}\right \}\right \}$ Maple : cpu = 0.009 (sec), leaf count = 16

$\left \{ y \left ( x \right ) =2\,{\frac {x}{{x}^{2}+2\,{\it \_C1}}} \right \}$

Hand solution

\begin {align} xy^{\prime }+xy^{2}-y & =0\nonumber \\ y^{\prime } & =\frac {1}{x}y-y^{2}\tag {1} \end {align}

This is of the form $$y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}$$ with $$f_{0}=0,f_{1}=\frac {1}{x},f_{2}=-1$$. Since $$f_{0}=0$$ this is Bernoulli diﬀerential equation. We always start by dividing by $$y^{2}$$$\frac {y^{\prime }}{y^{2}}=\frac {1}{x}\frac {1}{y}-1$ Then $$u=\frac {1}{y}$$ or $$y=\frac {1}{u}$$, therefore $$y^{\prime }=-\frac {u^{\prime }}{u^{2}}$$. Equating this to RHS of (1) gives\begin {align*} -\frac {u^{\prime }}{u^{2}}u^{2} & =\frac {1}{x}u-1\\ -u^{\prime } & =\frac {u}{x}-1\\ u^{\prime }+\frac {u}{x} & =1 \end {align*}

Integrating factor is $$e^{\int \frac {1}{x}dx}=x$$ and the above becomes$d\left ( xu\right ) =x$ Integrating\begin {align*} xu & =\frac {x^{2}}{2}+C\\ u & =\frac {x}{2}+\frac {C}{x}\\ & =\frac {x^{2}+2C}{2x} \end {align*}

Hence \begin {align*} y & =\frac {1}{u}\\ & =\frac {2x}{x^{2}+2C} \end {align*}

Veriﬁcation

restart;
ode:=x*diff(y(x),x)+x*y(x)^2-y=0;
my_solution:=2*x/(x^2+2*_C1);
odetest(y(x)=my_solution,ode);
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