Thread: Ode .. (2)

1. Ode .. (2)

Hello
I need help with the following ODE :

$\displaystyle \dfrac{dx}{(1-xy)^2}+\left[ y^2+\dfrac{x^2}{(1-xy)^2} \right]dy=0$

I multiplied both sides by $\displaystyle (1-xy)^2$, to get:

$\displaystyle dx+\left[ y^2(1-xy)^2+x^2 \right]dy=0$

Here, I tried the substitution $\displaystyle t=xy$. But it does not work, doesn't it?

Any help is appreciated .

2. If you write your ODE as

$\displaystyle \dfrac{dx}{dy} + (1-xy)^2y^2+x^2=0$
then this equation is Ricatti. One solution is $\displaystyle x = \dfrac{1}{y}.$

If you let $\displaystyle x = \dfrac{1}{y} + \dfrac{1}{u},$ the equation becomes linear.

3. Thanks.
But how did you know this substitution will make it linear?
By your experience? or there is an method to it?

4. Ricatti's equation has de form $\displaystyle y'(x)+P(x)y^2+Q(x)y=R(x),$ then if $\displaystyle y_1(x)$ is a particular solution, then by substituting $\displaystyle y(x)=y_1(x)+t$ you get a Bernoulli equation, but someone just did $\displaystyle y(x)=y_1(x)+\dfrac1t$ and that became the ODE into a linear one.