Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$

Question

Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$

This looks like a problem that would do well with a conversion to polar coordinates ($r^2 = x^2 + y^2$ and $x = r\cos(\theta)$ $y = r\sin(\theta)$). However, I am confused on how to change $y' = \frac{dy}{dx}$ to $\frac{dr}{d\theta}$, in order to do separation of variables. Could someone demonstrate a solution using this approach?

Answer 1

hint

Instead of looking for a cartesian solution of the form $ y=y(x) $, you want polar solution as $ r=r(\theta) $ with

$$x=r(\theta)\cos(\theta)\text{ and } y=r(\theta)\sin(\theta)$$

So, differentiation gives

$$dx=\Bigl(r'\cos(\theta)-r(\theta)\sin(\theta)\Bigr)d\theta$$

$$dy=\Bigl(r'\sin(\theta)+r(\theta)\cos(\theta)\Bigr)d\theta$$

thus, the equation becomes

$$\frac{dy}{dx}=\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}$$ $$=\frac{r-r\cos(\theta)}{r\sin(\theta)}$$ $$=\tan(\frac{\theta}{2})$$

which simplifies to

$$\frac{r'}{r}=-\frac{\cos(\frac{\theta}{2})}{\sin(\frac{\theta}{2})}$$

Yielding to the solution

$$\ln(\frac{r}{\lambda})=-2\ln(|\sin(\frac{\theta}{2})|)$$ and $$\boxed{r=\frac{2\lambda}{1-\cos(\theta)}}$$

Answer 2

$x = r \cos\theta, y = r \sin\theta$

$dx = \cos\theta \ dr - r \sin\theta \ d\theta$

$dy = \sin\theta \ dr + r \cos\theta \ d\theta$

Given equation is $dy = \frac{\sqrt{x^2+y^2}-x}{y} dx$

So we get,

$r \sin\theta \ (\sin\theta \ dr + r \cos\theta \ d\theta) = r \ (1 - \cos\theta) \ (\cos\theta \ dr - r \sin\theta \ d\theta)$

On simplifying we get,

$\displaystyle \frac{1}{r} \ dr = - \frac{\sin\theta}{1-\cos\theta} \ d\theta$

To integrate, simply observe that $1 - \cos\theta = t \implies \sin\theta \ d\theta = dt$

So both sides are simple integration.

Answer 3

$$y'=f(x,y)$$ $$\dfrac {dy}{dx}=f(x,y)$$ Apply the chain rule: $$\dfrac {dy}{d\theta}\dfrac {d\theta}{dx}=f(x,y)$$ $$\dfrac {dy}{d\theta}=g(r,\theta)\dfrac {dx}{d\theta}$$ Nowit's easy to compute the derivatives. $$\dfrac {dy}{d\theta}=\dfrac d {d\theta}(r(\theta)\sin (\theta))$$ $$\dfrac {dx}{d\theta}=\dfrac d {d\theta}(r(\theta)\cos (\theta))$$