How to solve a Riccati DE: $x^2(xy'+y)=2x^4+y^2$

Question

I need to find the general solution of this DE: $x^2(xy'+y)=2x^4+y^2$

I've changed it into this form: $y'=\frac{y^2}{x^3}-\frac{y}{x}+2x$, but I do not know how to proceed from here.

I'd appreciate general instructions on how to solve any Riccati DE.

Answer 1

There is no general solution to the Riccati equation. There are two main methods for solving the special forms of the Riccati equation. The first method and more powerful/complicated is to use Lie symmetries. The second method but less powerful/complicated method is the following.

Guess a solution of the type $y_1(x) = ax^n$. It is pretty easy to see that $n=2$, the factor $a$ can then be quickly determined by plugging $y_1(x)=ax^2$ into the ordinary differential equation (ODE).

Then use the substitution

$$y(x) = \dfrac{1}{u(x)} + y_1(x)$$ to reduce the ODE.

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You can also go back to the beginning of your calculations.

$$x^2(xy'+y)=2x^4+y^2$$ $$x^2(xy)'=2x^4+y^2$$ $$(xy)'=2x^2+(y/x)^2$$

Now substitute $y=z(x)/x$:

$$z' =2x^2+z^2/x^4.$$

Maybe now guessing the solution to this ODE is a little bit simpler.

Answer 2

write $$y'(x)=2x+\frac{y(x)^2}{x}-\frac{y(x)}{x}$$ and substitute $$y(x)=x^2u(x)$$ then we get $$u'(x)=\frac{u(x)^2-3u(x)+2}{x}$$ integrate both sides $$-\log(-u(x)+1)+\log(-u(x)+2)=\log(x)+C_1$$ now you can finish! the solution is $$y(x)=\frac{x^2(C_1x-2)}{C_1x-1}$$

Answer 3

$$x^2(xy'+y)=2x^4+y^2$$ It's clear that $y=x^2$ is a solution. $$x^2(x2x+x^2)=2x^4+x^4$$ The general solution should be of the form $y=vx^2$

$$x^2(x(v'x^2+2vx)+vx^2)=2x^4+v^2x^4$$ $$v'x+2v+v=2+v^2$$ $$v'x=2-3v+v^2$$ $$v'x=(v-1)(v-2)$$ It's seperable $$\int \frac {dv}{(v-1)(v-2)}=\ln(x)+K$$ $$\ln \frac {(v-2)}{(v-1)}=\ln(x)+K$$
$$ \boxed {y(x)=x^2\frac {(2-Kx)}{(1-Kx)}}$$