solving diff equation w/ bournoulli's equation

**snoboarder2k6** · February 14th 2008, 04:53 PM

I am trying to find the general solution of
$y(dy/dx) + x = sqrt(x^2 + y^2)<br />$

I'm having issues figuring out what n should be, as I can't get the y's together to figure out the powers. How would I do this.

thanks

**mr fantastic** · February 14th 2008, 05:58 PM

Originally Posted by snoboarder2k6

I am trying to find the general solution of
$y(dy/dx) + x = sqrt(x^2 + y^2)<br />$

I'm having issues figuring out what n should be, as I can't get the y's together to figure out the powers. How would I do this.

thanks

$y\frac{dy}{dx} + x = \sqrt{x^2 + y^2}$

$\Rightarrow \frac{dy}{dx} + \frac{x}{y} = \frac{\sqrt{x^2 + y^2}}{y}$

$\Rightarrow \frac{dy}{dx} + \frac{x}{y} = \sqrt{\frac{x^2 + y^2}{y^2}}$

$\Rightarrow \frac{dy}{dx} + \frac{x}{y} = \sqrt{\left( \frac{x}{y} \right)^2 + 1}$

The standard technique is to now make the substitution $y = vx \Rightarrow \frac{dy}{dx} = v + \frac{dv}{dx}$ . After some simplifying you get

$\frac{dv}{dx} + 1 + v^2 = \sqrt{1 + v^2}$ .

This DE is seperable, leading to $x = \frac{\sqrt{v^2 + 1} + 1}{v} + C$ . Now substitute back $y = vx \Rightarrow v = \frac{y}{x}$ .....

**snoboarder2k6** · February 14th 2008, 06:06 PM

ah, tricky, I forgot that's how to get something into a sgrt. thanks

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