0
$\begingroup$

I came across this diff eq that I tried solving using methods like u=y/x substitution but couldn't reduce it to anything meaningful. I am only familiar with solving first-order linear equations therefore could anyone suggest a starting point to approach this equation? Or perhaps known substitutions that I could use in order to reduce to a simpler form? Thanks.

$$\frac{y^2}{1-y^2}(\frac{x}{y} - y')^2 = 4$$

initial conditions y(0)=1

(mb for asking this question previously without any context)

Edit: First I simplified the equation:

$$\frac{x - yy'}{\sqrt{1-y^2}} = 2 ....(i)$$

I was able to solve the equation using the substitution $$u^2 = 1-y^2$$ with which I found

$$\frac{dy}{dx}= \frac{-u}{\sqrt{1-u^2}}\frac{du}{dx}$$

After substituting the relevant values of y and y' into eq(i),

$$\frac{x}{u}+u' = 2 ......(ii)$$

here is I realized that $$u=x .....(iii)$$ is a solution and therefore,

$$y=\sqrt{1-x^2}$$

would be the final solution, which checks out, but there is obviously a lack of rigour between eq(ii) and eq(iii), how can this step be reasoned mathematically?

$\endgroup$
3
  • $\begingroup$ Please edit your post to include your try, because maybe we can dissect it and see how you can proceed. $\endgroup$ Commented May 12, 2024 at 19:42
  • $\begingroup$ how about $u=y^2$? $\endgroup$ Commented May 12, 2024 at 19:44
  • $\begingroup$ thanks for the suggestion, after giving $u=y^2$ a shot, and then $u= 1-y^2$, i realized that $u^2 = 1-y^2$ is the best substitution, with which I could solve the equation. $\endgroup$ Commented May 12, 2024 at 21:47

1 Answer 1

Reset to default
1
$\begingroup$

Upon letting $u = \sqrt(1-y^2)$ , one ends-up with the "simpler" non-linear ODE: \begin{equation} x = 4u-u\frac{du}{dx} \end{equation} This ODE completely separates by making the substitution $u = xy$ , yielding standard integrals: \begin{equation} -\int\frac{dx}{x} = \int\frac{y.dy}{y^2-4y+1} \end{equation}

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.