# Differential Equation - Change of Variable

## Homework Statement

Find the general solution of

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

## 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 $$\frac{du}{-2u - \sqrt{u^2 + 1}}$$

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Mark44
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## Homework Statement

Find the general solution of

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

## 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.