Ricatti differential equation, there is one given solution but can't find Bernoulli equation.

Question

$$x^2y' + xy + x^2y^2 = 4$$

In this equation there are given one solution and this is

$$y=\frac{a}{x}$$

I think this is a Riccati equation and started solving by

$$y = v + \frac{a}{x}$$

and making derivation

$$y' = v' - \frac{a}{x^2}$$

So I ended up with this equation

$$v' = -v^2 -\frac{-v(2a+1)}{x} + \frac{4-a^2)}{x^2}$$

but I think this isn't a Bernoulli equation.

Is my starting point is wrong? Where can I go from this equation to find the general solution?

Answer 1

Hint: Substituting $$y(x)=\frac{v(x)}{x}$$ then we get $$\frac{dv(x)}{dx}=\frac{-v(x)^2+4}{x}$$ Can you proceed?

Answer 2

Hint: With $y=\dfrac1v+\dfrac{a}{x}$ you find $$v'+\dfrac3xv=1$$ which is a simple first order differential equation.

Answer 3

The first thing you should do is to insert the class of proposed solutions to determine the constant $a$. In general there is only a finite number of such solutions. You should end up with $a^2=4$, $a=\pm 2$.

Choosing $a=2$, your last equation reduces to the Bernoulli form $$v' = -v^2 +\frac{5v}{x}.$$

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Additionally to the other proposed solution strategies, you could directly insert the parametrization $y=\frac{u'}u$ to get $$ x^2u''+xu'=4u $$ which is a Cauchy-Euler equation with basis solutions $x^2$ and $x^{-2}$, thus $$ y(x)=\frac{2ax-2bx^{-3}}{ax^2+bx^{-2}}=\frac{2}x\cdot\frac{ax^4-b}{ax^4+b}. $$