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Differential Equation - Change of Variable

  • Thread starter cse63146
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  • #1
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Homework Statement



Find the general solution of

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


Homework Equations





The Attempt at a Solution



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

xu' + u = - u - [tex]\sqrt{u^2+1}[/tex]

u' = [tex]\frac{-2u - \sqrt{u^2 + 1}}{x}[/tex]

[tex]\frac{du}{-2u - \sqrt{u^2 + 1}} = \frac{dx}{x}[/tex]

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

Answers and Replies

  • #2
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4,838

Homework Statement



Find the general solution of

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


Homework Equations





The Attempt at a Solution



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

xu' + u = - u - [tex]\sqrt{u^2+1}[/tex]

u' = [tex]\frac{-2u - \sqrt{u^2 + 1}}{x}[/tex]

[tex]\frac{du}{-2u - \sqrt{u^2 + 1}} = \frac{dx}{x}[/tex]

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

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

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