Differential Equation - Change of Variable

[cse63146]

Homework Statement

Find the general solution of

y' = - \frac{x}{y} - \sqrt{(\frac{x}{y})^2 + 1}

Homework Equations

The Attempt at a Solution

I let y = ux -> y' + xu' + u

xu' + u = - u - \sqrt{u^2+1}

u' = \frac{-2u - \sqrt{u^2 + 1}}{x}

\frac{du}{-2u - \sqrt{u^2 + 1}} = \frac{dx}{x}

now I'm supposed to integrate both sides, just not sure how to find the integral of \frac{du}{-2u - \sqrt{u^2 + 1}}

 

[Mark44]

Homework Statement

Find the general solution of

y' = - \frac{x}{y} - \sqrt{(\frac{x}{y})^2 + 1}

Homework Equations

The Attempt at a Solution

I let y = ux -> y' + xu' + u

xu' + u = - u - \sqrt{u^2+1}

u' = \frac{-2u - \sqrt{u^2 + 1}}{x}

\frac{du}{-2u - \sqrt{u^2 + 1}} = \frac{dx}{x}

now I'm supposed to integrate both sides, just not sure how to find the integral of \frac{du}{-2u - \sqrt{u^2 + 1}}

From your substitution, y = ux, it follows that u = y/x. You replaced x/y by u, instead of 1/u.

Using the same substitution, I get
\frac{-u du}{1 + u^2 + \sqrt{1 + u^2}} = \frac{dx}{x}

That's still pretty ugly on the left side, but it might be amenable to completing the square in the denominator.