#### 2.108   ODE No. 108

$x y'(x)+y(x)+y(x)^2 (-\log (x))=0$ Mathematica : cpu = 0.0656555 (sec), leaf count = 15

$\left \{\left \{y(x)\to \frac {1}{c_1 x+\log (x)+1}\right \}\right \}$ Maple : cpu = 0.01 (sec), leaf count = 13

$\left \{ y \left ( x \right ) = \left ( 1+{\it \_C1}\,x+\ln \left ( x \right ) \right ) ^{-1} \right \}$

Hand solution

$$xy^{\prime }+axy^{2}+2y+bx=0$$This is Riccati non-linear ﬁrst order. Converting it to standard form\begin {align} xy^{\prime }-y^{2}\ln x+y & =0\tag {1}\\ y^{\prime } & =-\frac {1}{x}y+y^{2}\frac {\ln x}{x}\nonumber \\ & =f_{0}+f_{1}y+f_{2}y^{2}\nonumber \end {align}

This is Bernoulli non-linear ﬁrst order ODE since $$f_{0}=0$$. Dividing by $$y^{2}$$ gives

$\frac {y^{\prime }}{y^{2}}=-\frac {1}{x}\frac {1}{y}+\frac {\ln x}{x}$

Let $$u=\frac {1}{y}$$, hence $$u^{\prime }=-\frac {y^{\prime }}{y^{2}}$$, and the above becomes

\begin {align*} -u^{\prime } & =-\frac {1}{x}u+\frac {\ln x}{x}\\ u^{\prime }-\frac {1}{x}u & =-\frac {\ln x}{x} \end {align*}

Integrating factor is $$\mu =e^{\int -\frac {1}{x}dx}=e^{-\ln x}=\frac {1}{x}$$, hence

$d\left ( \mu u\right ) =-\mu \frac {\ln x}{x}$

Integrating

\begin {align*} \frac {1}{x}u & =-\int \frac {1}{x^{2}}\ln xdx+C\\ & =-\left ( -\frac {\ln x}{x}-\frac {1}{x}\right ) +C \end {align*}

Therefore

$u=\ln x+1+Cx$

Since $$u=\frac {1}{y}$$ then

$y=\frac {1}{\ln x+1+Cx}$

Veriﬁcation

restart;
ode:=x*diff(y(x),x)-y(x)^2*ln(x)+y(x)=0;
my_solution:=1/(ln(x)+1+_C1*x);
odetest(y(x)=my_solution,ode);
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