Solving differential equation using Riccati equation transformation I would like to request some help with the following problem. I need to solve this equation: $$y' +\frac{y}{x}+ xy^2 = 0, \space y(1)=1$$
using the Riccati equation, and the following transformation of said equation: 
$$u(x)=exp(\int h(x)y(x)dx)$$ 
and $$u''+(g(x)-\frac{h'(x)}{h(x)})u'(x)-k(x)h(x)u=0$$
I have shown that a solution to the original problem will be given in the form $\frac{log(u)'}{h(x)}$, but I am unable to find $u$, because every equation I arrive at somehow still contains $y$, which leaves me unable to solve the problem. How do I find the solution to the transformed version of the problem? Thank you in advance.
 A: Hint: Write $$-\frac{\frac{dy(x)}{dx}}{y(x)^2}-\frac{1}{xy(x)}=x$$ anf then sbstitute
$$v(x)=\frac{1}{y(x)}$$ and we get
$$\frac{dv(x)}{dx}-\frac{v(x)}{x}=x$$
with the integrating factor $$\mu(x)=\frac{1}{x}$$ we obtain
$$\frac{\frac{dv(x)}{dx}}{x}-\frac{v(x)}{x^2}=1$$
A: $$y' +\frac{y}{x}+ xy^2=0$$
Obviously they are several methods to solve this equation which is not a general Riccati equation but a degenerate form of Riccati equation. Nevertheless it is a Riccati equation and the OP is allowed to ask for solving it with a method especially devoted to the Riccati equations, even if the solving can be done with a simpler method not specific to Riccati equations.
In case of the general form of Piccati equation, when a particular solution cannot be guessed, the usual change in order to get the related second order linear equation is :
$$y=\frac{u'}{x\:u}$$
The coefficient of $u$ in the denominator is the coefficient of $y^2$ in the Riccati equation written on above form. See Eqs.(4) and (5) in http://mathworld.wolfram.com/RiccatiDifferentialEquation.html
$$y'=\frac{u''}{x\:u}-\frac{u'}{x^2\:u}-\frac{(u')^2}{x\:u^2}$$
Putting it into the original equation leads to :
$$\frac{u''}{x\:u}-\frac{u'}{x^2\:u}-\frac{(u')^2}{x\:u^2} +\frac{1}{x}\left(\frac{u'}{x\:u}\right)+ x\left(\frac{u'}{x\:u}\right)^2=0$$
After simplification :
$$u''=0$$
$$u=c_1x+c_2$$
$$y=\frac{u'}{x\:u}=\frac{c_1}{x(c_1x+c_2)}$$
$$y=\frac{1}{x(x+c_0)}\quad\text{with}\quad c_0=c_2/c_1$$
The condition $\quad y(1)=1=\frac{1}{1+c_0}\quad$ gives $c_0=0$. So the result is :
$$y(x)=\frac{1}{x^2}$$
A: Why Riccatti since it's separable ?
$$y' +\frac{y}{x}+ xy^2 = 0, \space y(1)=1$$
$$xy' +y+ x^2y^2 = 0$$
$$(xy)'+ (xy)^2 = 0$$
This equation is separable
$$\int \frac {d(xy)}{(xy)^2}=-\int dx$$
$$\frac 1{xy}=x+K$$
$$  y(x)=\frac 1 {x(x+K)}$$
A: By comparing coefficients in $y=\frac{u'}{hu}$,
$$
\frac{u''}{hu}-\frac{h'u'}{h^2u}-\frac{u'^2}{hu^2}+\frac{u'}{xhu}+\frac{xu'^2}{h^2u^2}=0\implies u''+\left[\frac{1}{x}-\frac{h'}{h}\right]u'+\frac{x-h}{h}\frac{u'^2}{u}=0,
$$ 
to eliminate the quadratic terms you should easily find $h=x$. Inserting this, $y=\frac{u'}{xu}$, and fixing the initial conditions as $u(1)=u'(1)=1$ leads to
$$
y'=\frac{u''}{xu}-\frac{u'}{x^2u}-\frac{u'^2}{xu^2}\implies \frac{u''}{xu}=0.
$$
This has the solution $u=x$, that is, $y=\frac1{x^2}$
