Solve the Differential equation $\frac{dy}{dx}=\frac{1}{x^4}+y^2$

Question

Solve the Differential equation $$\frac{dy}{dx}=\frac{1}{x^4}+y^2$$

My Try: Let $x^2y=t$ we have

$$y=\frac{t}{x^2}$$ Then

$$\frac{dy}{dx}=\frac{x^2dt-2txdx}{x^4}$$

But $$\frac{dy}{dx}=\frac{1+t^2}{x^4}$$ $\implies$

$$x^2dt-2txdx=1+t^2$$

any way to proceed from here?

Answer 1

$$\frac{dy}{dx}=\frac{1}{x^4}+y^2$$ This a Riccati's ODE. Applying the usual method to solve it (see note 2):

The change of function : $y(x)=-\frac{u'(x)}{u(x)}\quad;\quad y'=-\frac{u''}{u}+\frac{(u')^2}{u^2}$ transforms the ODE to :

$-\frac{u''}{u}+\frac{(u')^2}{u^2}=\frac{1}{x^4}+\left(-\frac{u'(x)}{u(x)}\right)^2$. After simplification : $$u''+\frac{1}{x^4}u=0$$ $$u=c_1x\sin\left(\frac{1}{x}\right)+c_2x\cos\left(\frac{1}{x}\right)$$ $u'=\left(c_1+\frac{c_2}{x}\right)\sin\left(\frac{1}{x}\right)+\left(c_2-\frac{c_1}{x}\right)\cos\left(\frac{1}{x}\right)$ $$y(x)=-\frac{\left(c_1+\frac{c_2}{x}\right)\sin\left(\frac{1}{x}\right)+\left(c_2-\frac{c_1}{x}\right)\cos\left(\frac{1}{x}\right)}{c_1x\sin\left(\frac{1}{x}\right)+c_2x\cos\left(\frac{1}{x}\right)}$$ Note 1 : This can be written with one arbitrary constant only $\frac{c_1}{c_2}$, or alternatively $\frac{c_2}{c_1}$.

Note 2 : Eqs.(4-6) in http://mathworld.wolfram.com/RiccatiDifferentialEquation.html

This transforms the Riccati ODE into a linear second order ODE. At first sight, increasing the order seems counterproductive. But it is often an efficient method because solving a linear second order ODE is generally simpler than solving a non-linear first order ODE. The linearity is the key property.

Answer 2

Substituting $$y(x)=-\frac{\frac{dv(x)}{dx}}{v(x)}$$ $$-\frac{\frac{d^2v(x)}{dx^2}}{v(x)}+\frac{\left(\frac{dv(x)}{dx}\right)^2}{v(x)^2}=\frac{\left(\frac{dv(x)}{dx}\right)^2}{v(x)^2}+\frac{1}{x^4}$$ $$-\frac{d^2v(x)}{dx^2}=\frac{v(x)}{x^4}$$ Let $$v(x)=xu(x)$$ then $$x\frac{d^2u(x)}{dx^2}+2\frac{du(x)}{dx}+\frac{u(x)}{x^3}=0$$ Let $t=\frac{1}{x}$ then $$\frac{\frac{d^2u(x)}{dx^2}}{t}+2\frac{du(x)}{dx}+t^3u(x)=0$$ then by the cahin rule $$-2t^2\frac{du(t)}{dt}+\frac{t^4\frac{d^2u(t)}{dt^2}+2t^3\frac{du(t)}{dt}}{t}+t^3u(t)=0$$ $$t^3\left(\frac{d^2u(t)}{dt^2}+u(t)\right)=0$$ this gives $$\frac{d^2u(t)}{dt^2}+u(t)=0$$ Solution: $$u(t)=C_1e^{it}+C_2e^{-it}$$