Solving $\frac{dy}{dx}=\frac{x^2y^2+xy+1}{x^2}$

Question

Assuming a substitution $v=xy$, isn't a standard prescription however it can help solving some ODEs such as $$\frac{dy}{dx}=\frac{x^2y^2+xy+1}{x^2}.$$ Let us use $xy=v \implies xy'+y=v'$, we get $$\frac{dv}{dx}=\frac{(v+1)^2}{x} \implies \frac{dv}{(v+1)^2}=\frac{dx}{x}.$$ Integrating both sides we get $$\frac{-1}{v+1}=\log x+C \implies y=-\frac{C+\log x+1}{x(C+\log x)}.$$

The question is: How else it can be solved

Edit: Unfortunately there was a typo of sign in the question, it should have been $+xy$ instead of $-xy$. It is corrected now. Any other method is most welcome.

Answer 1

$$\frac{\mathrm dy}{\mathrm dx}=\frac{x^2y^2-xy+1}{x^2}$$$$\implies x^2\mathrm dy = x^2y^2\mathrm dx - xy\mathrm dx +\mathrm dx$$ $$\implies x(x\mathrm dy + y\mathrm dx) = ((xy)^2+1)\mathrm dx$$ $$\implies \frac {\mathrm dx}{x}= \frac{\mathrm d(xy)}{(xy)^2 +1}$$

Update:

There was a typo in original post by OP. The concerned DE now is :

$$\frac{ \mathrm dy}{\mathrm dx}=\frac{x^2y^2+xy+1}{x^2}$$$$\implies x^3 \mathrm d \frac{y}{x} = (xy)^2\mathrm dx + \mathrm dx$$

$$\implies x\mathrm d (xy)-2yx\mathrm dx = (xy)^2\mathrm dx + \mathrm dx$$ where the last equation is obvious from the following substitution :

$$x^2 \mathrm d \frac{y}{x} = \mathrm d (xy) - 2y\mathrm dx$$

Now, it becomes the following variable separable form which is easy to solve. $$\implies \frac{\mathrm dx}{x} = \frac{\mathrm d (xy)}{(xy+1)^2}$$

Answer 2

Using $xy=v$ is a very good manner to solve the ODE. But one must not make a mistake :

$xy'+y=v'$ doesn't implies $\frac{dv}{dx}=\frac{(v-1)^2}{x}$.

It implies $\quad \frac{dv}{dx}=\frac{v^2+1}{x}\quad\implies\quad v=\tan(\ln|x|+C)\quad\implies\quad y=\frac{\tan(\ln|x|+C)}{x}$

Answer 3

As $y$ only occurs in the second degree, you can also consider this as Riccati equation. One solution method transforms it into a linear second order DE by setting $y=-\frac{u'}{u}$ to get $$ y'-y^2 = -\frac{u''}{u}=-\frac{u'}{xu}+\frac1{x^2}\\~\\ 0=x^2u''-xu'+u. $$ This now is an Euler-Cauchy equation with characteristic polynomial $r(r-1)-r+1=(r-1)^2$, so its basis solutions are $x$ and $x\ln x$, $$ u=x(A+B\ln x),~~~u'=A+B(1+\ln x)\\~\\ \implies y=-\frac{A+B(1+\ln x)}{x(A+B\ln x)} $$ Now you can cancel either $A=BC$ or $B=AC$, each variant containing one solution the other does not.

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In the original form of the question this works too, then the Euler-Cauchy equation $$ 0=x^2u''+xu'+u $$ has the characteristic polynomial $r(r-1)+r+1=r^2+1$ leading to basis solutions $x^{\pm i}$ or the real pair $\cos(\ln x),\sin(\ln x)$ so that $$ u=A\cos(\ln x)+B\sin(\ln x),~~u'=\frac1x(-A\sin(\ln x)+B\cos(\ln x))\\~\\ \implies y=\frac1x\,\frac{A\sin(\ln x)-B\cos(\ln x)}{A\cos(\ln x)+B\sin(\ln x)} =\frac1x\,\frac{A\tan(\ln x)-B}{A+B\tan(\ln x)} $$

Answer 4

You have gotten two nice analytical approaches.

I will give a more mechanical and numerical approach that is more general when analytical approaches fail.

We will use the substitution $v = xy$

which by Jacquelin gives $$\frac{d v}{dx} = \frac{v^2+1} x$$

We will now assume a power series expansion around $0$.

Using some well known facts

1. The square of a power series expansion is a convolution of it's coefficients.
2. Differentiation and multiplication and division with x are linear operators (- but only for as long as we can ensure we never need to represent reciprocals of x and their powers). Let us call the corresponding operators D and X.

using the notation $v_k$ is the k:th iteration to find a solution we can define a fixed point iteration

$$v_k = D^{-1}X^{-1}(v_{k-1} ^ 2 + 1)$$

Running a loop in Gnu Octave for a power series expansion truncated to a 16 order polynomial gives me a solution with first four terms very close to [0,1,0.5,1/3]. Which I find reasonable as power series expansion of $-log(1-t)$ starts like that and $t\to\tan(t)$ is close to $t\to t$ when $t$ is close to $0$.

The following terms start differing from the $t^{k}/k$ pattern $\left[\begin{array}{llll}0&1&0.5&0.333333333333333\\0.229166666666667&0.158333333333333&0.109490740740741&0.0757275132275133\\0.0523773354828043&0.0362272728664511&0.0250569590651333&0.0173309023741605\\0.0119870966097813&0.00829099853635585&0.00573455433506161&0.0039663634345758\end{array}\right] $

Perhaps someone can help me see which C it corresponds to?

Answer 5

$$\frac{dy}{dx}=\frac{x^2y^2+xy+1}{x^2}.$$ Riccati's DE can also be solved by inspection Try $y_p=\dfrac ax$ then: $$\dfrac {-a}{x^2}=\dfrac {a^2+a+1}{x^2}$$ $$(a+1)^2=0$$ $$ \implies a=-1$$ $$y_p=-\dfrac 1x$$ Now the DE can be transformed into a Bernoulli's DE: Substitute: $$y=-\dfrac 1 x +u$$ Another way to see this substitution is to rewrite the original DE by completing the square on RHS: $$\frac{dy}{dx}=\dfrac{x^2y^2+xy+1}{x^2}.$$ $$\frac{dy}{dx}=\dfrac{x^2y^2+\color{red}{2xy}+1}{x^2}-\dfrac yx$$ $$\frac{dy}{dx}\color{red}{-\dfrac 1 {x^2}}=\dfrac{(xy+1)^2}{x^2}-\dfrac yx\color{red}{-\dfrac 1 {x^2}}$$ $$\left(y+\frac{1}{x}\right)'=\left(y+\dfrac 1x\right)^2-\dfrac 1 {x}\left(y+\frac{1}{x}\right)$$ This is Bernoulli's differential equation.