Riccati Differential Equation Help Show that $y(x)=\frac{1}{x}$ is a solution of the Ricatti differential equation:
$$y'+\frac{1-3xy}{x^2}+3y^2=0$$
Hence using a suitable transformation of the dependant variable, solve the differential equation when $y(1)=3$.
Any help appreciated.
 A: Multiply by $x^2$ and substitute $y=Q\frac{w'}{w}$
$$x^2Q'\frac{w'}{w}+x^2Q\frac{w''}{w}-x^2Q\frac{(w')^2}{w^2}+1-3xQ\frac{w'}{w}+3x^2Q^2\frac{(w')^2}{w^2}=0$$
We choose what $Q$ is so we choose it conveniently:
$$-x^2Q\frac{(w')^2}{w^2}+3x^2Q^2\frac{(w')^2}{w^2}=0$$
We get $Q=\frac 13$. Now the equation reads:
$$\frac13x^2\frac{w''}{w}+1-x\frac{w'}{w}=0$$
Multiply by $w$
$$x^2 w''(x)-3 x w'(x)+3 w(x)=0$$
This is a Cauchy Euler equation. Try the solution $x^r$ (I am assuming you can do this)
The solutions are $w=c_2 x^3+c_1 x$
Bring it all together:
$$y=\frac 13\frac{3c_2x^2+c_1}{c_2 x^3+c_1 x}$$
Divide the num and denum by $c_2$ and call the new constant $c$
$$y=\frac{3x^2+c}{ 3x^3+3c x}$$
And the final solution:
$$\frac{4 x^2-1}{x \left(4 x^2-3\right)}$$
Your answer is wrong. It is a solution to the equation, but it does not satisfy the initial condition.
A: here is another way to solve this problem. we will make a change of variable $$v = \frac1{xy}, y =\frac1{xv}, y' = -\frac1{x^2v}-\frac{v'}{xv^2}\tag 1 $$
subbing $(1)$ in $$y'+\frac{1-3xy}{x^2}+3y^2=0, $$ we find that 
$$-\frac1{x^2v}-\frac{v'}{xv^2} +\frac{1-3/v}{x^2}+\frac3{x^2v^2} = 0$$ multiplying by $x^2v^2$ gets us $$ -v-xv'+v^2-3v+3=0\to xv' = (v-1)(v-3)$$
leading to $$\int_{1/3}^v \frac{dv}{(v-1)(v-3)}= \frac12\int_{1/3}^v \frac{dv}{v-3} - \frac{dv}{v-1} = \int_1^x\frac{dx}x = \ln (x) $$
that is $$\ln\left(\frac{3-v}{4(1-v)}\right) = 2\ln x \to \frac{3-v}{1-v} =4x^2\to v = \frac{3-4x^2}{1-4x^2}$$ and finally $$y = \frac{1-4x^2}{x(3-4x^2)}.$$
A: Differentiate $\frac{1}{x}$ and substitute in the differential equation. Then the general solution is $y\frac{c}{x}$.Using $y(1)=3$ you get that $c=3$. Thus the general solution is $y=\frac{3}{x}$
